[
    "### Question:\nYou are given a 0-indexed permutation of n integers nums.\nA permutation is called semi-ordered if the first number equals 1 and the last number equals n. You can perform the below operation as many times as you want until you make nums a semi-ordered permutation:\n\nPick two adjacent elements in nums, then swap them.\n\nReturn the minimum number of operations to make nums a semi-ordered permutation.\nA permutation is a sequence of integers from 1 to n of length n containing each number exactly once.\n \nExample 1:\n\nInput: nums = [2,1,4,3]\nOutput: 2\nExplanation: We can make the permutation semi-ordered using these sequence of operations: \n1 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3].\n2 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4].\nIt can be proved that there is no sequence of less than two operations that make nums a semi-ordered permutation. \n\nExample 2:\n\nInput: nums = [2,4,1,3]\nOutput: 3\nExplanation: We can make the permutation semi-ordered using these sequence of operations:\n1 - swap i = 1 and j = 2. The permutation becomes [2,1,4,3].\n2 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3].\n3 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4].\nIt can be proved that there is no sequence of less than three operations that make nums a semi-ordered permutation.\n\nExample 3:\n\nInput: nums = [1,3,4,2,5]\nOutput: 0\nExplanation: The permutation is already a semi-ordered permutation.\n\n \nConstraints:\n\n2 <= nums.length == n <= 50\n1 <= nums[i] <= 50\nnums is a permutation.\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def semiOrderedPermutation(self, nums: List[int]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the minimum number of swaps of adjacent elements to turn a permutation into a semi-ordered one. A semi-ordered permutation has 1 at the first position and n at the last. Let me think about how to approach this.\n\nFirst, let me understand the problem. The permutation is a list of numbers from 1 to n, each exactly once. The goal is to get 1 to the beginning and n to the end with as few adjacent swaps as possible.\n\nLet me think about the steps. Let's say I first find where the 1 is located and where the n is. For example, if 1 is at position i and n (which is the length of the array) is at position j.\n\nMoving 1 to the first position requires i swaps. Because each swap moves it one position left. Like if it's at index 2, you need two swaps to get to index 0.\n\nSimilarly, moving n to the last position (index n-1) requires (n-1 - j) swaps. But wait, there's a catch. What if the position of n is before the position of 1? Because when we move 1 to the left, it might shift the position of n.\n\nFor example, let's say the original array is [3, 1, 4]. Here, n is 3. Let's say n is at index 0. Then 1 is at index 1. So moving 1 to index 0 takes 1 swap. Then moving n (which is 3) to the end... but after moving 1, the array becomes [1, 3, 4]. Now, n is at index 1. So moving it to index 2 requires one swap. Total swaps: 2.\n\nBut originally, if we first move n to the end, what would happen? Original positions: 3 is at 0, 1 at 1. Moving 3 to index 2 would require two swaps. Then 1 is already at index 0. So total swaps would be 2. Same result.\n\nBut here, the initial positions: let's say we first find the positions of 1 and n. Let's call them pos1 and posn. So moving 1 to the front takes pos1 steps. Moving n to the end takes (n-1 - posn) steps. However, if posn is before pos1, then when we move 1 to the left, the position of n would be shifted by one. Because when you move 1 left past the n's original position, the n's position increases by one. Wait, no. Let me think again.\n\nSuppose in the original array, n is at index j, and 1 is at index i, where j < i. So when you move 1 from i to 0, which takes i swaps. During these swaps, when the 1 moves past the n's original position (j), which was between 0 and i. When you move 1 left through j, the n (which was originally at j) would be shifted one position to the right. Therefore, when we first move 1 to the front, the position of n increases by 1 if it was originally before the 1.\n\nSo when we first move 1, then when moving n to the end, the original position of n may have changed.\n\nTherefore, the total steps for moving 1 and n depends on their initial positions. So the initial positions are pos1 and posn. Let me calculate:\n\nIf posn > pos1: Then moving 1 first reduces the position of 1 by pos1 steps. Then moving n requires (n - 1 - posn) steps. Because since n was after 1's original position, moving 1 to the front doesn't affect n's position.\n\nBut if posn < pos1: When moving 1 to the front, which takes pos1 steps. But during that process, the n is shifted right by 1, because when you move 1 past it. Therefore, the original posn is now at posn + 1 (since the 1 moved past it). Therefore, moving n to the end would require (n-1 - (posn + 0))? Wait, no. Let me think. Suppose originally, posn is j, and pos1 is i, with j < i. When you move 1 left to position 0, each swap moves it left by one. So during each swap, the 1 moves left, passing elements. Any element between the original position of 1 and the new position (0) will be shifted. For example, if the original position of 1 is i, then after moving to 0, all elements before i are shifted right by one. Wait no. When you swap adjacent elements to move left, each step shifts the 1 left by one and the element to the left moves right by one. So, if the n is at position j, and during moving the 1 from i to 0, if j is less than i, then during the movement, when the 1 passes through j's position, the n (which was at j) would be swapped once to the right. So overall, after moving 1 to the front, the position of n would be at j + 1 if it was originally before the 1's position.\n\nTherefore, in this case, when posn < pos1: moving 1 first will cause the position of n to be (original posn + 1). Therefore, moving n to the end requires (n-1 - (posn + 1 - (original?))) steps. So the total steps would be:\n\nsteps = pos1 (moving 1 to front) + (n-1 - posn - 1 + 1?) Let me think.\n\nLet me take an example. Example 2 from the problem: input [2,4,1,3]. Here, n is 4, and the array is length 4. pos1 is 2 (index starts at 0). posn is 1 (the value 4 is at index 1). So moving 1 to front: needs 2 steps (from index 2 to 0). But during that movement, since 4 is at index 1 (original posn=1 < pos1=2):\n\nAfter first swap (index 2 and 1): array becomes [2,1,4,3]. The 4 is now at index 2. Then after moving to index 0 (another swap with index 1 and 0?), no. Wait the movement of 1 to position 0 from original position 2 requires two swaps. So during that movement, the 1 moves past any elements in between. For example, in original array [2,4,1,3], moving 1 left once (swap index 1 and 2) gives [2,1,4,3]. Then swap index 0 and 1 to get [1,2,4,3]. So during this process, the 4 was at index 1. After moving 1 past it (to index 1), the 4 is at index 1 (no, in the first swap, when 1 moves to index 1, the 4 is swapped with 1, moving to index 2. Then when 1 moves left again to index 0, the 2 (at index 0) and 1 (index 1) are swapped, which doesn't involve the 4. So the final position of 4 is index 2. Original posn was 1. Now it's 2. So increased by 1. So in this case, after moving 1, the position of n is original posn + 1.\n\nTherefore, moving n from its new position (original posn +1) to end requires (n-1 - (original posn) ) steps. Because new position is original posn +1? Wait, original posn is j. new posn is j+1. So to get to end, steps needed are (n-1 - (j+1)) + 1? No, because moving from position j+1 to position n-1 requires (n-1 - (j+1)) swaps. Each adjacent swap moves it one position. So total steps for moving n is (n-1 - j -1) = n - j -2.\n\nSo total steps for moving 1 and n is pos1 (steps to move 1) + (n - j - 2) steps to move n. But j is original posn. So total steps would be pos1 + (n - posn - 2) = (pos1 - posn) + (n-2). But how?\n\nIn example 2, pos1 is 2 (original), posn is 1. So st",
    "### Question:\nYou are given a 0-indexed string s that consists of digits from 0 to 9.\nA string t is called a semi-repetitive if there is at most one consecutive pair of the same digits inside t. For example, 0010, 002020, 0123, 2002, and 54944 are semi-repetitive while 00101022, and 1101234883 are not.\nReturn the length of the longest semi-repetitive substring inside s.\nA substring is a contiguous non-empty sequence of characters within a string.\n \nExample 1:\n\nInput: s = \"52233\"\nOutput: 4\nExplanation: The longest semi-repetitive substring is \"5223\", which starts at i = 0 and ends at j = 3. \n\nExample 2:\n\nInput: s = \"5494\"\nOutput: 4\nExplanation: s is a semi-reptitive string, so the answer is 4.\n\nExample 3:\n\nInput: s = \"1111111\"\nOutput: 2\nExplanation: The longest semi-repetitive substring is \"11\", which starts at i = 0 and ends at j = 1.\n\n \nConstraints:\n\n1 <= s.length <= 50\n'0' <= s[i] <= '9'\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def longestSemiRepetitiveSubstring(self, s: str) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the longest semi-repetitive substring in a given string of digits. Let me think about what a semi-repetitive substring is. Oh right, it's a substring that has at most one consecutive pair of same digits. So like, if there are two pairs, then it's invalid. For example, \"0010\" is okay because there's one pair (the two 0s). But something like \"00110\" would have two pairs (the two 0s and two 1s) so it's not allowed.\n\nSo the task is to find the maximum length of such a substring in the given string. Then return that length.\n\nHmm. How do I approach this? Well, since the string is up to 50 characters long, maybe a brute-force approach is feasible. Because even O(n^3) would be 50*50*50 = 125,000 operations, which is manageable.\n\nSo here's an idea: iterate over all possible substrings (start and end indices), check for each substring whether it meets the semi-repetitive condition, and track the maximum length where it does.\n\nBut how to check if a substring meets the condition?\n\nFor a given substring, we need to check that there is at most one pair of consecutive same digits. So, in the substring, we can count the number of times s[i] == s[i+1]. If that count is <= 1, then it's valid.\n\nWait, but the substring is contiguous. So for example, in the substring \"5223\", there is exactly one pair (the two 2s). That's allowed. If there's a longer run, like \"222\", then the count would be two pairs (positions 0-1 and 1-2), so that's two pairs. Which would make it invalid. So the maximum allowed is one pair.\n\nSo for any substring, we can check consecutive duplicates. Let's think about how to count that. For a substring from i to j (inclusive), we need to check for each k from i to j-1, how many times s[k] == s[k+1]. If the total is <=1, then it's valid.\n\nSo for example, the substring \"5223\" has one such pair. The substring \"52233\" would have two pairs (positions 1-2 and 3-4), so that's two, which is invalid.\n\nSo the plan is:\n\n- For all possible substrings (i to j), check the number of consecutive duplicates. If it's <=1, track the maximum length.\n\nNow, how to implement this.\n\nFirst, for each possible start i and end j (where j >= i), we can process the substring s[i..j]. For each such substring, check the condition.\n\nBut how to efficiently check the condition? For each substring, we can loop through its characters and check consecutive pairs.\n\nBut since the string is small (max 50 characters), even with O(n^3) time, it's acceptable.\n\nLet me outline the steps:\n\nInitialize max_length to 1 (since the minimum substring is length 1, which has zero consecutive pairs).\n\nLoop over i from 0 to len(s)-1:\n\n   Loop over j from i+1 to len(s)-1:\n\n      current substring is s[i..j], which has length j - i + 1.\n\n      Now, check how many consecutive duplicates are there in this substring.\n\n      To check this, initialize count = 0.\n\n      For k from i to j-1:\n\n          if s[k] == s[k+1], increment count.\n\n          if count > 1, break early.\n\n      if count <= 1, then update max_length if current length is larger.\n\nSo for each possible substring, we check the count. If it's valid, consider its length.\n\nOnce all substrings are checked, return the max_length.\n\nYes, this seems manageable.\n\nLet me test this logic against the examples.\n\nExample 1: s = \"52233\"\n\nPossible substrings of length >=2. Let's see for the longest ones.\n\nThe entire string is length 5. Let's check how many consecutive duplicates. Let's see:\n\nOriginal string is 5,2,2,3,3.\n\nBetween indexes 0-4:\n\nCheck positions 0-1: 5 vs 2: no.\n\nPositions 1-2: 2-2: yes. count=1.\n\nPositions 2-3: 2 vs 3: no.\n\nPositions 3-4: 3-3: yes. count=2. So total is 2, which is more than 1. So the entire string is invalid.\n\nSo the maximum valid substring could be up to 4 characters. Let's check \"5223\" (indexes 0-3):\n\nConsecutive pairs:\n\n0-1: no.\n\n1-2: yes. count=1.\n\n2-3: 2 vs 3: no. Total count is 1. So valid. Hence length 4. Which matches the example.\n\nAnother example is \"5494\" (example 2). The entire string has no consecutive duplicates. So the maximum is 4.\n\nFor example 3, \"1111111\", the entire string has multiple consecutive pairs. So the maximum valid substring is 2 (exactly one pair allowed?), but wait: the maximum allowed is at most one consecutive pair. So for a substring of length 2 (\"11\"), there is exactly one pair. For longer substrings, like \"111\" (length 3), there are two consecutive pairs (positions 0-1 and 1-2). So that's two, which is invalid. Hence the maximum is 2.\n\nSo the logic works here.\n\nNow, let's think about edge cases.\n\nAnother example: s = \"11\". The substring is the entire string, which has one consecutive pair. So valid. Length 2. So answer is 2.\n\nIf s is \"111\", then the maximum valid substring would be 2. Either the first two 1's or last two.\n\nAnother case: s is \"1212121\". The entire string has no consecutive duplicates, so the answer is 7.\n\nAnother case: s is \"112211\". The entire string has two pairs (positions 0-1 and 2-3 and 4-5). So when checking the entire string, the count is 3, which is invalid. But substrings like 0-3 (\"1122\") has two pairs (1-1 and 2-2), which is invalid. So what substring is valid? For example, \"112\" would have two consecutive 1s and then 1 and 2. Count is 1. So valid. But longer ones?\n\nLike, for example, \"11221\" would have two pairs (positions 0-1, 2-3). So count is 2. invalid. So the maximum for \"112211\" would be 4? For example, the substring from 0-3 (1122) is invalid. But if you take from 0-2 (112, count 1). Or maybe take 1-5 (12211) which would have two pairs. Hmm. Not sure. Maybe the maximum is 4 when there's a substring that has exactly one pair.\n\nBut perhaps there is a substring of length 4 that has exactly one pair. For example, \"1221\" (positions 1-4): check consecutive duplicates. 1 and 2 (no), 2 and 2 (yes, count 1), then 2 and 1 (no). So count is 1. So that's valid. Length 4.\n\nSo yes. The maximum would be 4 in that case.\n\nSo the algorithm would pick that.\n\nNow, how to implement this in Python.\n\nThe starter code is a class Solution with a function longestSemiRepetitiveSubstring that takes s as input.\n\nLet me think about the code.\n\nFirst, get the length of s. Let's say n = len(s). For n=1, the answer is 1 (but since the constraints say 1<=s.length, but according to the problem statement, the substring is non-empty. However, since in the examples, for n=1, there's no consecutive pairs, so it's allowed. So the maximum possible is 1. But according to the problem statement, the examples for input \"5\" would return 1.)\n\nBut according to the problem's sample inputs, let's see. For example, in example 3, input is \"1111111\", output is 2. But for a substring of length 1, it's allowed but the max length would be 1. But since 2 is larger, the output is 2. So for n=1, the answer is 1.\n\nSo our code needs to handle n=1 as well.\n\nBut back to the algorithm.\n\nInitialize max_len as 1, since any single character is allowed (but if the string is longer, maybe there's a pair",
    "### Question:\nYou are given a 0-indexed integer array nums and an integer threshold.\nFind the length of the longest subarray of nums starting at index l and ending at index r (0 <= l <= r < nums.length) that satisfies the following conditions:\n\nnums[l] % 2 == 0\nFor all indices i in the range [l, r - 1], nums[i] % 2 != nums[i + 1] % 2\nFor all indices i in the range [l, r], nums[i] <= threshold\n\nReturn an integer denoting the length of the longest such subarray.\nNote: A subarray is a contiguous non-empty sequence of elements within an array.\n \nExample 1:\n\nInput: nums = [3,2,5,4], threshold = 5\nOutput: 3\nExplanation: In this example, we can select the subarray that starts at l = 1 and ends at r = 3 => [2,5,4]. This subarray satisfies the conditions.\nHence, the answer is the length of the subarray, 3. We can show that 3 is the maximum possible achievable length.\nExample 2:\n\nInput: nums = [1,2], threshold = 2\nOutput: 1\nExplanation: In this example, we can select the subarray that starts at l = 1 and ends at r = 1 => [2]. \nIt satisfies all the conditions and we can show that 1 is the maximum possible achievable length.\n\nExample 3:\n\nInput: nums = [2,3,4,5], threshold = 4\nOutput: 3\nExplanation: In this example, we can select the subarray that starts at l = 0 and ends at r = 2 => [2,3,4]. \nIt satisfies all the conditions.\nHence, the answer is the length of the subarray, 3. We can show that 3 is the maximum possible achievable length.\n\n \nConstraints:\n\n1 <= nums.length <= 100 \n1 <= nums[i] <= 100 \n1 <= threshold <= 100\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def longestAlternatingSubarray(self, nums: List[int], threshold: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the longest subarray in nums that starts with an even number, alternates even and odd between consecutive elements, and all elements in the subarray are <= threshold. Let me think about how to approach this.\n\nFirst, let me understand the conditions again. The subarray must start at l, end at r. The first condition is nums[l] must be even. Then, for all i between l and r-1, the current element's parity must alternate with the next. So like even, odd, even, odd... or odd, even, etc. But since the first element must be even, the sequence would start with even, then alternate. Also, every element in the subarray must be <= threshold.\n\nSo the steps I need to take are:\n\n1. Check each possible starting point l where nums[l] is even and nums[l] <= threshold. For each such l, find the maximum possible r such that the conditions are satisfied.\n\n2. For each starting l, expand as far right as possible while maintaining the conditions.\n\nHow can I efficiently check this? Since the constraints are up to 100 elements, a brute-force approach might be acceptable. Because even O(n^2) would be 100*100=10,000 steps, which is manageable.\n\nSo here's a plan:\n\nInitialize max_length to 0.\n\nIterate over each possible starting index l from 0 to len(nums)-1:\n\n- If nums[l] is even and <= threshold, then proceed. Else, skip this l.\n\n- For this l, check each possible end index r starting from l up to len(nums)-1.\n\n- We need to check if from l to current r, all elements meet the parity alternation and each element is <= threshold.\n\nWait, but checking all r for each l might be time-consuming, but given the small constraints, it's acceptable.\n\nBut perhaps we can find a way to compute the maximum possible r for a given l more efficiently. Let's think.\n\nFor a given l (starting even and valid), the subarray must alternate parities between consecutive elements. So for position l, the next element (l+1) must be odd, then l+2 even, etc. So the parity alternates. Also, each element must be <= threshold.\n\nSo for a given l, we can try to extend the subarray as long as possible. Let's simulate this:\n\nStart at l. Check if it's even and <= threshold. If not, can't start here. If yes, then check next element.\n\nThen check the next element (l+1) must be odd, and <= threshold. If either condition fails, stop. Else, proceed. Then check l+2 must be even, and <= threshold. And so on.\n\nSo for each l, we can simulate this process and track the maximum length possible.\n\nThis way, for each l, we can check step by step how far we can go. Let's think about how to implement this.\n\nSo for each l in 0..n-1:\n\nif nums[l] is even and <= threshold:\n\n   current_max_length = 1 (since the subarray can start here, at least length 1)\n\n   if current_max_length is already better than global max, update.\n\n   then, check next element (i+1). For the next element:\n\n   for i from l to r (but maybe just step by step):\n\n      check if the next element's parity is different from current. Also check if it's <= threshold.\n\nSo for example, for l=1 in example 1: [3,2,5,4], threshold 5.\n\nWait example 1: nums is [3,2,5,4]. The subarray starts at l=1, which is 2. Then next element is 5 (odd), which is okay. Also, 5 <=5 is yes. Then next element (4) must be even. Also, 4 <=5. So the subarray is 2,5,4. So length 3.\n\nSo during the simulation for l=1, we check:\n\nstart with 2, current_max_length=1.\n\nCheck index 1+1=2. nums[2]=5. Its parity is odd (since previous was even). Also 5<=5. So current_max_length becomes 2.\n\nCheck index 3. Previous parity was odd (at index 2), so next (index 3) must be even. 4 is even. And 4 <=5. So current_max_length becomes 3. So for l=1, the max length is 3.\n\nWhich is the answer.\n\nSo for each l, once we start, we can iterate through the array from l onwards, checking each next element.\n\nThus, the steps are:\n\nfor l in 0 to len(nums)-1:\n\n   if nums[l] is even and <= threshold:\n\n       current_length = 1\n\n       if current_length > max_len: update max_len\n\n       for r in l+1 to len(nums)-1:\n\n           check if nums[r] <= threshold? If not, break, since can't include this or previous.\n\n           also check if the parity alternates between previous element (nums[r-1]) and current (nums[r]).\n\n           So for the current r, check if nums[r] % 2 != nums[r-1] % 2.\n\n           If both conditions are met, then current_length +=1, check if it's the new max. Else, break.\n\nSo once any condition fails for a particular r, we can't extend further for this l, so break.\n\nWait, but how do I check the parity alternation? For example, in the sequence, each consecutive pair must alternate. So when adding a new element at r, I just need to check if nums[r] has a different parity than nums[r-1]. Because if that's true for every consecutive pair, then the entire subarray alternates.\n\nYes. For example, if the first element is even, then the next must be odd (so even and odd), then the next must be even (so odd and even), etc. So each consecutive pair alternates.\n\nTherefore, for a given l, once we check each next element (starting from l+1), we can check for each step whether the current element meets the parity condition and threshold. If at any step it fails, then the subarray can't be extended further. So for l, the maximum possible is up to r-1.\n\nTherefore, the algorithm would be:\n\nInitialize max_len = 0.\n\nFor each l from 0 to n-1:\n\n   if nums[l] is even and nums[l] <= threshold:\n\n       current_length = 1\n\n       max_len = max(max_len, current_length)\n\n       for r in l+1 to n-1:\n\n           if nums[r] > threshold: break\n\n           if (nums[r] % 2) == (nums[r-1] % 2): break\n\n           current_length +=1\n\n           max_len = max(max_len, current_length)\n\n   # else, do nothing, since the subarray can't start here.\n\nSo for example, in example 3: nums = [2,3,4,5], threshold 4.\n\nCheck l=0: nums[0] is even and 2 <=4. So proceed.\n\ncurrent_length is 1. max is 1.\n\nr=1 (nums[1] is 3). Check threshold: 3 <=4. Yes. Check parity: 3 is odd, previous (nums[0] is even). Different. So okay. current_length becomes 2. max is 2.\n\nr=2: nums[2]=4. Check threshold: 4 <=4. Yes. Check parity: previous (nums[1] is odd), current even. Different. So okay. current_length 3. max is 3.\n\nr=3: nums[3] is 5. Check threshold: 5 >4. Break. So for l=0, the maximum is 3. Which is the correct ",
    "### Question:\nYou are given a 0-indexed array usageLimits of length n.\nYour task is to create groups using numbers from 0 to n - 1, ensuring that each number, i, is used no more than usageLimits[i] times in total across all groups. You must also satisfy the following conditions:\n\nEach group must consist of distinct numbers, meaning that no duplicate numbers are allowed within a single group.\nEach group (except the first one) must have a length strictly greater than the previous group.\n\nReturn an integer denoting the maximum number of groups you can create while satisfying these conditions.\n \nExample 1:\n\nInput: usageLimits = [1,2,5]\nOutput: 3\nExplanation: In this example, we can use 0 at most once, 1 at most twice, and 2 at most five times.\nOne way of creating the maximum number of groups while satisfying the conditions is: \nGroup 1 contains the number [2].\nGroup 2 contains the numbers [1,2].\nGroup 3 contains the numbers [0,1,2]. \nIt can be shown that the maximum number of groups is 3. \nSo, the output is 3. \nExample 2:\n\nInput: usageLimits = [2,1,2]\nOutput: 2\nExplanation: In this example, we can use 0 at most twice, 1 at most once, and 2 at most twice.\nOne way of creating the maximum number of groups while satisfying the conditions is:\nGroup 1 contains the number [0].\nGroup 2 contains the numbers [1,2].\nIt can be shown that the maximum number of groups is 2.\nSo, the output is 2. \n\nExample 3:\n\nInput: usageLimits = [1,1]\nOutput: 1\nExplanation: In this example, we can use both 0 and 1 at most once.\nOne way of creating the maximum number of groups while satisfying the conditions is:\nGroup 1 contains the number [0].\nIt can be shown that the maximum number of groups is 1.\nSo, the output is 1. \n\n \nConstraints:\n\n1 <= usageLimits.length <= 10^5\n1 <= usageLimits[i] <= 10^9\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def maxIncreasingGroups(self, usageLimits: List[int]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the maximum number of groups that can be formed under certain conditions. Let me try to understand the problem again.\n\nThe problem says that each group must consist of distinct numbers. So in a single group, each element can be used once. Also, each subsequent group must have a length strictly greater than the previous. For example, if the first group has size 1, the next must be at least 2, then the next 3, etc. So the total required \"usage\" for k groups is like 1 + 2 + 3 + ... +k = k*(k+1)/2. But how does this relate to the usageLimits given?\n\nBut here's the thing: each element i can be used up to usageLimits[i] times across all groups. So, for each group of size m (say the j-th group), we need m distinct elements. And the total number of times each element is used across all groups can't exceed their limit.\n\nThe goal is to find the maximum k such that we can form k groups with sizes strictly increasing (group 1 size 1, group 2 size 2, ..., group k size k?), or maybe not exactly that. Wait, the problem says each group after the first must be longer than the previous. So if there are k groups, the sizes must form a strictly increasing sequence. The maximum possible k is what we need to find. However, the sizes could be any strictly increasing sequence. But to maximize k, the optimal way would be to have the sizes as 1, 2, 3,...,k. Because that's the minimal possible total sum for k groups. For example, if you have group sizes 1, 2, 3, that's sum 6. If they were 2,3, sum is 5 but k is 2. So for maximum k, we need the minimal total usage, which is 1+2+...+k = k*(k+1)/2.\n\nTherefore, the problem reduces to finding the maximum k where the sum of the first k terms (1+2+...+k) can be covered by the usageLimits. But how?\n\nBut how do we check if a given k is feasible? Let's think.\n\nEach group of size m requires m distinct elements. For the total usage across all groups, each element's contribution must be <= their usage limit. Let's think about how much each element contributes. For example, if we have k groups, each of size 1, 2,...,k, then each element is part of some number of groups. For example, elements in the first group (size 1) are used once. Those in the second (size 2) are used twice, etc. Wait, no. Each element can be in multiple groups. Wait, but how? Wait, the problem doesn't say that once an element is used in a group, it can't be reused in another group. The only constraint is that for each element, the total number of groups it's part of is <= its usage limit. Oh right! Because the problem says that each number can be used across groups as long as the total is within the limit.\n\nSo the problem becomes: arrange the elements into groups, strictly increasing sizes, and assign elements to groups such that each element's count across groups is within their limit. The maximum k.\n\nBut how can we model this? For a given k, how do we check if it's possible?\n\nLet me think about the approach. Since each group of size m requires m distinct elements, the total number of \"slots\" needed across all groups is sum_{i=1}^k i = S = k(k+1)/2. However, each element can contribute a certain number of slots (equal to how many groups it's part of). For example, an element with a usage limit of 3 can be in up to 3 different groups, each time contributing to a different slot.\n\nBut how do we arrange the elements so that their total contribution meets the required S, and also for each group, the elements in it are distinct?\n\nWait, perhaps the key is that for each group m (assuming group m has size m), we need m distinct elements. However, elements can be reused in different groups. But the constraints are:\n\n- For each element i, the number of groups it's in is ≤ usageLimits[i].\n\nBut how to arrange them optimally to reach maximum k?\n\nAn idea comes to mind: sort the usageLimits in non-decreasing order. Then, for each possible k, check if the sum of the usageLimits is sufficient to meet the required sum S, and also that for each step up to k, the cumulative usage is enough.\n\nBut even if the total sum is sufficient, the arrangement may not be possible. For example, if all usage limits are 1. The total sum is n. For k, the maximum possible is 1, since 1+2=3 exceeds sum if n is 2. But even if sum is sufficient, how to check?\n\nWait, let's take example 3. Input is [1,1]. The sum is 2. For k=2, the required sum is 3, which is larger than 2, so k=1 is the maximum. Which matches the example.\n\nAnother example: usageLimits sorted as [1,2,5]. Let's think for k=3. The required sum is 3*4/2=6. The sum of the usageLimits is 1+2+5=8 ≥ 6. So possible. But how?\n\nBut even if the total sum is sufficient, we need to arrange the elements such that for each group size 1, 2, 3. How?\n\nLet me think of a way to check if a given k is feasible. Let's sort the usageLimits in non-decreasing order. Then, for each i-th group (when trying to build up to k groups), we need to have enough elements. \n\nWait, here's an idea inspired by greedy approach. For each m from 1 to k, the first m groups require a total of 1+2+...+m elements. But each element can contribute up to its usage limit. However, for the m-th group (which has size m), we need m new elements. But since the groups are increasing, each element can be part of multiple groups. However, perhaps for each m-th step (when trying to form the first m groups), we need that the sum of the smallest m usageLimits is >= the sum 1+2+...+m. Or perhaps not. \n\nAlternatively, here's a thought: when trying to form k groups, each group i (for i from 1 to k) has size i. For each element, it can contribute to multiple groups. However, when building up the groups, for each 1<=m<=k, the total number of elements that have usage count >=1, >=2, etc. ?\n\nAnother approach is inspired by the problem of finding the maximum number of groups in a sorted array. For example, when you have sorted usage limits, for each position, you can check how many elements are sufficient for a certain group count.\n\nLet me think of the following method. Sort the usageLimits in non-decr",
    "### Question:\nYou are given two 0-indexed integer arrays nums1 and nums2 of equal length. Every second, for all indices 0 <= i < nums1.length, value of nums1[i] is incremented by nums2[i]. After this is done, you can do the following operation:\n\nChoose an index 0 <= i < nums1.length and make nums1[i] = 0.\n\nYou are also given an integer x.\nReturn the minimum time in which you can make the sum of all elements of nums1 to be less than or equal to x, or -1 if this is not possible.\n \nExample 1:\n\nInput: nums1 = [1,2,3], nums2 = [1,2,3], x = 4\nOutput: 3\nExplanation: \nFor the 1st second, we apply the operation on i = 0. Therefore nums1 = [0,2+2,3+3] = [0,4,6]. \nFor the 2nd second, we apply the operation on i = 1. Therefore nums1 = [0+1,0,6+3] = [1,0,9]. \nFor the 3rd second, we apply the operation on i = 2. Therefore nums1 = [1+1,0+2,0] = [2,2,0]. \nNow sum of nums1 = 4. It can be shown that these operations are optimal, so we return 3.\n\n\nExample 2:\n\nInput: nums1 = [1,2,3], nums2 = [3,3,3], x = 4\nOutput: -1\nExplanation: It can be shown that the sum of nums1 will always be greater than x, no matter which operations are performed.\n\n \nConstraints:\n\n1 <= nums1.length <= 10^3\n1 <= nums1[i] <= 10^3\n0 <= nums2[i] <= 10^3\nnums1.length == nums2.length\n0 <= x <= 10^6\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to solve this problem where I have two arrays nums1 and nums2, and I need to find the minimum time to make their sum (of nums1) less than or equal to x. The process is that every second, each element in nums1 increases by the corresponding element in nums2. Then, after that, I can perform an operation to set any one element of nums1 to zero. The challenge is to figure out the optimal way to perform these operations over time to minimize the total time required.\n\nFirst, I need to understand how the sum evolves over time. Let's think about what happens each second. Suppose we don't perform any operations. Then, after t seconds, each element in nums1 would have increased by nums2[i] multiplied by t. But since each second we can set one element to zero, we need to decide which elements to reset at which times to minimize the total sum.\n\nHmm. Let me think about the problem differently. For each element i, if we decide to reset it k times (but here, each time we can reset only once, probably?), but no, the problem allows us to choose any index each second. Wait, but each second we can perform one operation. So over t seconds, we can perform t operations, each time resetting a certain index. But since there are n elements, maybe the optimal strategy is to reset some elements multiple times. However, given that each time you reset an element, it sets it to zero. But then, in subsequent steps, that element will start accumulating again. So perhaps it's better to reset elements with higher nums2[i], because they contribute more over time. \n\nBut how do we model this? Let's consider that each element can be reset at most once. Because if you reset it more than once, say twice, then there's some time between the first and second reset. But maybe resetting it once at the optimal time is better than resetting it twice. For example, if you reset it once at time t, then again at time t', but perhaps during that time it started accumulating again. So maybe resetting each element once is sufficient. Because once you reset it, you've removed the accumulated value up to that point. But if you reset it again later, you might be losing some accumulated growth. \n\nWait, but the operation is done after each second's increment. So for example, during each second, all elements are increased by their nums2[i], then you can perform the operation. So if you reset an element at time t, then during the next seconds, it starts increasing again. Therefore, maybe the optimal strategy is to reset each element exactly once, at some optimal time. Because resetting it again would just be redundant unless new accumulation happened. \n\nSo perhaps the problem reduces to selecting for some elements the time at which we reset them, and for each element, the contribution to the total sum depends on when we reset it. \n\nLet me formalize this. Let's say for element i, we decide to reset it once at time t_i. Then, the contribution of element i to the total sum would be:\n\nThe initial value of nums1[i] is a_i, and nums2[i] is b_i. \n\nBefore the first reset, during each second, the element accumulates. When we reset it at time t_i, we eliminate all the accumulated value up to that point. However, after that, it starts accumulating again. Wait no, if we reset it once, then after that, it will accumulate again. But if we don't reset it again, then during the entire process, after time t_i, it starts accumulating again. \n\nBut if we are trying to minimize the total sum, perhaps it's better to reset elements with higher b_i as late as possible? Or as early as possible? Wait, let's think. Suppose we reset element i at time t. Then, during the first t seconds, it's accumulating. But when you reset it, you set it to zero. However, during those t seconds, before the reset, it was accumulating. Wait, no. Because the operation is done after the increment. So for example, if you reset the element at time t, then during each of the t seconds, the element is increased by b_i each time. But when you reset it, you set it to zero. \n\nWait, let me think step by step. Let's take element i. Initially, it's a_i. \n\nAt each time step s (from 1 to t), the following happens:\n\n- All elements are increased by their b_i. Then, you can perform an operation. \n\nIf you decide to reset element i at time s, then after the increment, you set it to zero. \n\nSo for the first s-1 times, the element was increased by b_i each time. Then, during the s-th step, it's increased again (so after that, it's a_i + b_i * s). Then you reset it to zero. \n\nAfter that, during the next steps (from s+1 to t), the element is increased by b_i each time. But since it was reset, it starts from zero. \n\nBut if we are considering the total sum at time t, then for element i, if it was reset once at time s (during step s), then during the steps after s, it's being increased by b_i each step until time t. \n\nSo the total contribution of element i to the sum at time t would be:\n\nIf it was never reset during the t steps: a_i + b_i * t. \n\nIf it was reset once at step s (so during step s, after the increment, it's set to zero). Then during steps s+1 to t, it accumulates b_i*(t - s). So the total contribution would be the value before reset (a_i + b_i*s) is set to zero, but then for (t - s) steps, it's added by b_i each time. So the total contribution is 0 + b_i*(t - s) ?\n\nWait, no. Because during each step after s, the element is increased by b_i. So if we reset it at step s, then during step s, after the increment and operation, it's zero. Then, during step s+1, before the operation, it is increased by b_i again. So after step s, the element is zero. Then during step s+1, it's increased by b_i, then perhaps reset again or not. \n\nSo, for example, if we reset element i once at step s, then during the entire process up to time t, the total contribution of element i is:\n\nThe amount accumulated up to step s (a_i + b_i*s) is set to zero. Then, from step s+1 to step t, it accumulates (t - s) times. So the value is 0 + b_i * (t - s). Wait, but that's only if we don't reset it again. So if we reset it once, then during the remaining steps, it's accumulating again. \n\nTherefore, if we reset element i once at time s, then its contribution to the total sum at time t is b_i*(t - s). \n\nBut what if we reset it multiple times? For example, if we reset it at s1 and s2 (s1 < s2). Then during step s1, after increment, it's set to zero. Then during step s2, after increment (which adds b_i*(s2 - s1) since last reset?), no. Wait, during each step, the element is increased by b_i. So after being reset to zero at step s1",
    "### Question:\nYou are given an integer n representing the number of houses on a number line, numbered from 0 to n - 1.\nAdditionally, you are given a 2D integer array offers where offers[i] = [start_i, end_i, gold_i], indicating that i^th buyer wants to buy all the houses from start_i to end_i for gold_i amount of gold.\nAs a salesman, your goal is to maximize your earnings by strategically selecting and selling houses to buyers.\nReturn the maximum amount of gold you can earn.\nNote that different buyers can't buy the same house, and some houses may remain unsold.\n \nExample 1:\n\nInput: n = 5, offers = [[0,0,1],[0,2,2],[1,3,2]]\nOutput: 3\nExplanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers.\nWe sell houses in the range [0,0] to 1^st buyer for 1 gold and houses in the range [1,3] to 3^rd buyer for 2 golds.\nIt can be proven that 3 is the maximum amount of gold we can achieve.\n\nExample 2:\n\nInput: n = 5, offers = [[0,0,1],[0,2,10],[1,3,2]]\nOutput: 10\nExplanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers.\nWe sell houses in the range [0,2] to 2^nd buyer for 10 golds.\nIt can be proven that 10 is the maximum amount of gold we can achieve.\n\n \nConstraints:\n\n1 <= n <= 10^5\n1 <= offers.length <= 10^5\noffers[i].length == 3\n0 <= start_i <= end_i <= n - 1\n1 <= gold_i <= 10^3\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def maximizeTheProfit(self, n: int, offers: List[List[int]]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. The problem is about maximizing the profit from selling houses to buyers. Each buyer wants a range of houses, and once you accept an offer, you can't use those houses again. So the goal is to select non-overlapping offers such that the total gold is maximized. \n\nHmm, this reminds me of the interval scheduling problem. Oh right, like the classic activity selection problem where you choose the maximum number of non-overlapping intervals. But here, instead of counting the number, each interval has a value (gold), and we need to maximize the total value. So this is similar to the weighted interval scheduling problem.\n\nRight, in the weighted interval scheduling problem, you sort the intervals by end time and use dynamic programming. Let me think how to approach this here.\n\nFirst, I need to process the offers in a way that allows me to make optimal choices. Let's consider sorting all the offers based on their end_i. Then, for each offer, we can determine the maximum profit up to that point by either taking the offer (and adding its gold to the maximum profit up to the previous non-overlapping offer) or not taking it (so the max profit remains as before).\n\nSo the steps would be:\n\n1. Sort all offers by their end_i in ascending order.\n2. For each offer, compute the maximum profit considering up to that offer.\n\nBut how do I handle multiple offers? Let's think about dynamic programming. Let's say dp[i] represents the maximum profit considering the first i offers (assuming the offers are sorted). Wait, but maybe a better way is to have a DP array where dp[i] represents the maximum profit for the first i houses. Hmm, but with n up to 1e5, that might not be efficient. Or maybe not. Let's think again.\n\nAlternatively, since the offers can be considered as intervals, when sorted by end points, for each offer, we can find the latest offer that does not conflict with it. For example, for an offer j that ends at end_j, the previous non-overlapping offer would be the one with end_i < start_j. So, for each offer, we can binary search in the sorted list to find that.\n\nBut how do I structure this in Python?\n\nLet me outline the steps again:\n\n- Sort all offers by their end_i (increasing order).\n- For each offer i, find the maximum profit if we take this offer. To take it, we need to add its gold to the maximum profit up to the earliest end that is before (start_i + 1?), or before start_i? Wait, the offers are for ranges of houses. Since once you take an offer, all houses in that offer's start_i to end_i are used. So, for a new offer, if we take it, we need to find the last offer that ends before the start of the current offer's start. For example, if the current offer is [s, e, g], then any previous offer that ends before or at s-1 is okay.\n\nSo, for each offer in the sorted list, the profit if we take it is: the gold of this offer plus the maximum profit achievable up to the end of the previous offer which ends before s-1.\n\nSo how do I perform this efficiently?\n\nLet me think. Let's have a list of sorted offers by end. Then, for each offer i (with end e_i), we can perform a binary search on the sorted list to find the largest index j where the end of offer j is <= start_i -1. Then the DP[i] = max(DP[i-1], DP[j] + current offer's gold). Or maybe the DP array is built such that dp[i] is the maximum profit considering the first i offers. Wait, perhaps the DP approach here is to have dp[i] represent the maximum profit considering all offers up to the i-th (sorted by end). Then the answer would be dp[m-1], where m is the number of offers.\n\nAlternatively, another approach is to have a DP array where dp[i] represents the maximum profit considering houses up to i. For each house, we can decide whether to sell some offers that end at i or not. But how?\n\nWait, maybe the first approach (weighted interval scheduling) is more straightforward here. Let's proceed with that.\n\nLet me first sort the offers by their end. Let's say we have a list sorted_offers, sorted by end in increasing order.\n\nThen, for each offer i, we can compute the maximum profit up to i. Let's define dp[i] as the maximum profit considering the first i offers. But how do we find the previous compatible offer?\n\nAlternatively, for each offer in the sorted list, we can compute the maximum profit up to that offer. The maximum profit for the first i offers is either including this offer or not. If including, we need to find the last offer that ends before or at start_i -1, and add that's max profit plus current's gold. Otherwise, take the max profit up to i-1.\n\nSo, let's define dp[i] as the maximum profit considering the first i offers (after sorting). Then, for the i-th offer, when processing it, we can find the largest j where end_j <= start_i -1. Then dp[i] = max(dp[i-1], dp[j] + current gold).\n\nBut how to handle the indexes here? Let's think. For example, if the sorted offers are in an array, then for each offer in the array, we can have their end in increasing order.\n\nSo, for each offer in the sorted list, we can perform a binary search on the list to find the largest index j where end <= start_i -1.\n\nBut since the offers are sorted by end, we can use binary search here. For example, for the current offer's start, we can search for the largest end that is <= start -1. The corresponding index would be found via binary search.\n\nBut how to get that? Let's think of a helper function. For example, given a list of offers sorted by end, for a particular value x (start_i -1), find the largest index where end <= x.\n\nYes. So for each offer in the sorted list, during processing, we can perform this binary search on the list up to the current index.\n\nBut what about multiple offers with same end? The binary search would find the appropriate one.\n\nOnce that's done, the DP approach would be:\n\nInitialize dp[0] as the gold of the first offer (if there's no previous offer that ends before its start, then it's just the gold of the first offer, or 0? No, if there's no such previous offer, then we can take this offer, so the value is max of taking it or not taking it. Wait, but dp[i] is the maximum profit up to i-th offer. Let me think again.\n\nActually, the dp array here is for the sorted list of offers. So for example, dp[i] represents the maximum profit considering the first i offers (sorted by end). For each i-th offer (starting from 0?), we can choose to take it or not.\n\nBut how to compute dp[i]. The recurrence is:\n\ndp[i] = max(dp[i-1], offer_i's gold + dp[j]), where j is the last offer that ends before s",
    "### Question:\nYou are given two strings s and t of equal length n. You can perform the following operation on the string s:\n\nRemove a suffix of s of length l where 0 < l < n and append it at the start of s.\n\tFor example, let s = 'abcd' then in one operation you can remove the suffix 'cd' and append it in front of s making s = 'cdab'.\n\nYou are also given an integer k. Return the number of ways in which s can be transformed into t in exactly k operations.\nSince the answer can be large, return it modulo 10^9 + 7.\n \nExample 1:\n\nInput: s = \"abcd\", t = \"cdab\", k = 2\nOutput: 2\nExplanation: \nFirst way:\nIn first operation, choose suffix from index = 3, so resulting s = \"dabc\".\nIn second operation, choose suffix from index = 3, so resulting s = \"cdab\".\n\nSecond way:\nIn first operation, choose suffix from index = 1, so resulting s = \"bcda\".\nIn second operation, choose suffix from index = 1, so resulting s = \"cdab\".\n\nExample 2:\n\nInput: s = \"ababab\", t = \"ababab\", k = 1\nOutput: 2\nExplanation: \nFirst way:\nChoose suffix from index = 2, so resulting s = \"ababab\".\n\nSecond way:\nChoose suffix from index = 4, so resulting s = \"ababab\".\n\n \nConstraints:\n\n2 <= s.length <= 5 * 10^5\n1 <= k <= 10^15\ns.length == t.length\ns and t consist of only lowercase English alphabets.\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def numberOfWays(self, s: str, t: str, k: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the number of ways to transform string s into t in exactly k operations. The operation allowed is taking a suffix of length l (where 0 < l < n) and moving it to the front. And since k can be up to 1e15, I need an efficient way to compute this, probably using matrix exponentiation or some dynamic programming with fast exponentiation.\n\nFirst, let's understand the operation. For example, given a string s of length n, each operation corresponds to choosing a suffix length l (from 1 to n-1), cutting that suffix, and putting it in front. So for each operation, there are (n-1) possible choices. But not all choices will lead to the same string.\n\nThe problem is to find, after exactly k operations, how many different ways there are to reach t from s. \n\nLet me think about how the operations affect the string. Each operation is a permutation of the string. However, different choices of l will result in different permutations. But since the problem is about counting the number of paths (ways) that lead from s to t in exactly k steps, I need to model this as transitions between states.\n\nSo perhaps we can model this as a state transition problem. Let's consider each possible rotation (or permutation) that the string can be in. But since the actual string can be very long, maybe we can find equivalence classes based on the rotation. For example, each operation is a rotation by some l. Let's think about rotations in terms of how they shift the characters.\n\nFor example, rotating by l is equivalent to rotating by l mod n? Well, for example, if n=4 and l=2, rotating by 2 steps would give the same result as rotating by 2 + 4*m steps. Wait, but each operation is a rotation by l steps where l is between 1 and n-1. But each operation can be considered as adding a rotation. However, combining multiple operations can lead to various rotations.\n\nBut perhaps the key is to find for each possible rotation r (i.e., how much the string is rotated from the original), the number of ways to reach that rotation after k operations. Then, if the initial string is s, and after k operations it becomes a rotation of s, we can check if that rotation equals t.\n\nWait, but the problem is that each operation is a rotation. So, applying multiple operations leads to some composite rotation. However, rotations can be composed. Let me think: suppose I have two operations. The first operation is rotating by l1, then the second by l2. How does that combine?\n\nLet me think in terms of rotation steps. Rotating by l is equivalent to moving the last l characters to the front. For example, the original string is S = A B C D (n=4). Rotating by l=1 would take D and put it in front: D A B C. Rotating again by l=1 would take C and put it in front: C D A B. So each rotation by l is equivalent to a left rotation of (n - l) steps? Because rotating by l steps to the right (as in this operation) is the same as rotating left by (n - l) steps. For example, rotating right by 1 step is equivalent to rotating left by 3 steps. \n\nAlternatively, perhaps the total rotation after k operations can be modeled as some value modulo n. Let me think: each operation adds a rotation of l (in terms of right rotation). But how do multiple operations combine?\n\nLet me think of the rotation as a permutation. Suppose after one operation, the string is rotated by l1. Then another operation of l2. What is the total effect?\n\nLet me take an example. Original string is ABCD (positions 0-3). After rotating by l=1 (taking last 1 character and putting in front), it becomes DABC. Now, if I rotate this by l=2, taking the last 2 characters (BC) and putting them in front: BC DA → BCDA. Let's see what the total rotation is. The original string was ABCD. BCDA is a rotation. Let's see: ABCD → DABC (right rotation by 1). Then DABC → BCDA. How is BCDA a rotation of the original? Let's see:\n\nOriginal ABCD. BCDA is a rotation. Let me think of left rotations. Left rotation by 1: BCDA. Oh right! So the first operation was a right rotation by 1 (equivalent to left rotation by 3), then a right rotation by 2 (equivalent to left rotation by 2). How do these combine?\n\nLet me think of the rotation in terms of left rotations. Let me denote that each right rotation by l is equivalent to a left rotation by (n - l) mod n. \n\nLet me model the total rotation as cumulative left rotations. Suppose after each operation, the total rotation is updated. For example, if the current rotation is R (left rotation), and then we perform a right rotation by l steps (equivalent to left rotation by m = n - l steps), then the total rotation becomes (R + m) mod n. \n\nWait, but how do successive rotations work? Let's think of the string as being rotated. Suppose you have a current rotation of R steps to the left. Applying another left rotation of m steps would result in R + m steps. But if you apply a right rotation (which is equivalent to a left rotation of m = n - l steps), then the total rotation becomes R + m mod n. \n\nYes. So combining rotations can be modeled as adding the corresponding left rotation steps modulo n.\n\nSo, each operation (right rotation by l) contributes (n - l) steps to the total left rotation. \n\nBut each operation can choose any l between 1 and n-1. Therefore, for each operation, there are (n-1) possible choices. Each choice corresponds to adding a certain value (m) to the total rotation, where m ranges over certain values. For example, for each possible l (1 ≤ l ≤ n-1), m can be (n - l) mod n. Since l ranges from 1 to n-1, m ranges from 1 to n-1 as well. Because when l=1, m = n-1; when l=2, m = n-2; ... when l = n-1, m=1. So for each operation, the added left rotation can be any value from 1 to n-1. However, each value m in 1..n-1 can be achieved in exactly one way (since for m, l is n - m).\n\nBut wait, for each operation, the number of choices that add a specific m (from 1 to n-1) is exactly 1. Because each m corresponds to exactly one l. Therefore, for each operation, there are (n-1) possibilities, each contributing a different m (each m from 1 to n-1 once).\n\nTherefore, the problem reduces to: starting with a total rotation of 0 (original string), after k operations, what is the number of ways that the total rotation is equal to r (mod n), where r is such that rotating s by r left steps gives t. Then, the answer depends on finding the correct r and computing the number of ways to reach that r after k steps.\n\nBut first, we need to find all possible rotations r such that rotating s by r left steps gives t. Let me think: for example, if the original string is s, then after a left ro",
    "### Question:\nYou are given a 0-indexed array nums consisting of positive integers.\nThere are two types of operations that you can apply on the array any number of times:\n\nChoose two elements with equal values and delete them from the array.\nChoose three elements with equal values and delete them from the array.\n\nReturn the minimum number of operations required to make the array empty, or -1 if it is not possible.\n \nExample 1:\n\nInput: nums = [2,3,3,2,2,4,2,3,4]\nOutput: 4\nExplanation: We can apply the following operations to make the array empty:\n- Apply the first operation on the elements at indices 0 and 3. The resulting array is nums = [3,3,2,4,2,3,4].\n- Apply the first operation on the elements at indices 2 and 4. The resulting array is nums = [3,3,4,3,4].\n- Apply the second operation on the elements at indices 0, 1, and 3. The resulting array is nums = [4,4].\n- Apply the first operation on the elements at indices 0 and 1. The resulting array is nums = [].\nIt can be shown that we cannot make the array empty in less than 4 operations.\n\nExample 2:\n\nInput: nums = [2,1,2,2,3,3]\nOutput: -1\nExplanation: It is impossible to empty the array.\n\n \nConstraints:\n\n2 <= nums.length <= 10^5\n1 <= nums[i] <= 10^6\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def minOperations(self, nums: List[int]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. The problem is about finding the minimum number of operations to remove all elements from an array. The allowed operations are either removing two elements of the same value or removing three elements of the same value. And if it's not possible, return -1.\n\nHmm. So first, I need to figure out for each number how many times it occurs. Because each occurrence count will determine how we can split the operations. For example, if a number appears 4 times, how do we split that into operations of 2 or 3?\n\nWait, right. Let's think about each frequency. For each frequency, we need to find the minimum number of operations (steps) that can sum up to that frequency, using steps of 2 or 3. But also, if it's possible.\n\nWait, but each operation of 2 removes two elements. So for a count, how can we split it into groups of 2 and 3 such that the total number of groups (operations) is minimized?\n\nWait, but the problem requires that all elements are removed. So for each count, we need to check if it's possible to express it as a combination of 2s and 3s. And find the minimal number of operations needed.\n\nSo first, for each number's count, let's think about possible ways to split into 2 and 3. Let's take some examples.\n\nFor example, count is 1: impossible. Because you can't take 2 or 3 from 1. So if any number has a count of 1, return -1.\n\nCount 2: one operation of 2.\n\nCount 3: one operation of 3.\n\nCount 4: two operations of 2. Because 2*2=4. Or maybe another way? Like 3 + 1? No, but 1 isn't allowed. So two operations of 2.\n\nCount 5: Let's see. 2 + 3 = 5. So that's two operations. Which is better than, say, two 2s and one 1 (which isn't allowed). So two operations.\n\nWait, how about count 5: 2*2 + 1*3? No, but 1 isn't allowed. Wait, 2 + 3 is exactly 5, so two operations. So that's possible.\n\nCount 6: 3*2 operations of 3 (two operations, 3+3=6). Or three operations of 2 (3*2=6). Which is better? Well, we need the minimal number of operations. So for 6, two operations of 3 (each removes 3 elements) would be better than three operations of 2. So total 2 operations.\n\nSo for each count, we need to find the minimal number of operations. How do we compute this for each frequency?\n\nSo the problem reduces to, for each frequency f, find the minimal number of steps (operations) that can sum to f, where each step is either 2 or 3. If not possible, return -1.\n\nBut how to compute that?\n\nLet me think about how to model this. For a given frequency f, what's the minimal number of operations?\n\nWe need to find non-negative integers a and b such that 2a + 3b = f. Then the number of operations is a + b. Find the minimal a + b.\n\nAlternatively, since a and b are non-negative integers, perhaps there's a way to compute this without checking all possibilities.\n\nLet me think for a given f.\n\nWe can iterate over possible values of b (number of 3s) and see what's left, then check if the remainder can be divided by 2.\n\nBut how to find the minimal steps?\n\nAlternatively, for each f, the minimal steps can be found as follows:\n\nThe idea is that using as many 3s as possible may lead to minimal steps, but sometimes not. For example, if f is 4, using two 2s is better than one 3 and then 1 (which is invalid). So how?\n\nLet me think. For f, let's try all possible numbers of 3s that can be used. For example, for f, the maximum possible 3s is f//3. So for each possible b in 0 to max_b, we can check if (f - 3*b) is even and non-negative. If so, then (f -3*b)/2 is a, and total steps is a + b = (f-3b)/2 + b = (f + 3b)/2 ?\n\nWait, no. Let me recalculate. Let a be the number of 2's, and b be the number of 3's. Then 2a + 3b = f. The total steps is a + b. So solving for a: a = (f - 3b)/2. Since a must be non-negative integer.\n\nSo for each possible b (number of 3's) such that 3b <= f and (f - 3b) is even and >= 0, then compute a + b = (f -3b)/2 + b = (f -3b + 2b)/2 = (f -b)/2. Wait, but how can we find the minimal steps?\n\nAlternatively, since the total steps is a + b = (f -3b)/2 + b = f/2 - (3b)/2 + b = f/2 - b/2. So to minimize this, for a given f, we need to maximize b (the number of 3's) possible, but such that (f -3b) is even and non-negative.\n\nWait, for example, let's take f=5. The maximum possible b is 1 (since 3*1=3 <=5). Then f -3b is 5-3=2, which is even. So steps is 2/2 + 1 = 1 + 1 = 2. Which is correct.\n\nAnother example, f=4. The maximum b is 1. 3*1=3, f-3b=1 which is odd. So can't take. Try b=0. Then a=4/2=2 steps. So total steps is 2. Which is correct.\n\nAnother example: f=7. Let's see. Possible b's. Let's try b=2 (3*2=6, 7-6=1, invalid). b=1: 3, 7-3=4, even. So a=2. steps is 2+1=3. Alternatively, b=0: 7/2 is 3.5, no. So minimal steps is 3.\n\nAnother example: f=8. Let's think. What's the best way? 3*2 + 2*1 = 8. So two 3's and one 2. Total steps is 2+1=3. Or four 2's: steps 4. So better to take 3 steps. So how to compute this?\n\nSo the approach for each f is to find the maximum possible number of 3's. Let me think. How to find that?\n\nLet me think of f mod 3.\n\nCase 1: f mod 3 == 0. Then we can take all 3's. So steps is f/3. Which is optimal.\n\nCase 2: f mod 3 == 1. Then, perhaps we need to replace two 3's with three 2's. Because for example, if f is 4 (which is 3 +1). But 4 can't be 3 +1. Because 1 is left. So instead, 2*2's. But for f=4, that's two steps. Another example: 7. Which is 3*2 +1. So here, replace two 3's with three 2's (since 7 mod 3 is 1. So 7 = 3*2 +1. But we can't take that. So subtract 2*3=6, left with 1, no. So instead, perhaps subtract one 3 (so 7-3=4). Now, 4 can be split as two 2's. So total steps is 1 (for the 3) + 2 (for the two 2's) = 3 steps. Alternatively, perhaps adjust the number of 3's.)\n\nAlternatively, when f mod 3 is 1, then we can take (f-4)/3 3's and add one 2's. Because 4 is 2*2. For example, 4 is 4, which is 2*2. For 7, which is 3+4, so 1+ 2 steps. So total 3 steps. Similarly, 10: 10 mod 3 is 1. 10-4=6, which is 3*2. So 2+2 steps (for 6/3 is 2 steps plus 1 step for 4? No, 2 steps for 4 (two 2s). So total 2+2=4 steps. But 10 can be split as 3*2 + 2*2 = 6 +4, total 4 steps. Which is same as before.)\n\nCase 3: f mod 3 == 2. Then, just add one more 2. Because f is 3k +2. So steps is k + 1. For example, f=5 (3*1 +2). Steps is 1+1=2. Which is correct.\n\nSo how to compute the steps for each frequency:\n\nFor each frequency f:\n\nIf f is 1: impossible.\n\nElse:\n\nCompute how many 3's plus possible adjustments.\n\nLet me think of the formula:\n\nmax_triples = f // 3\n\nthen check f % 3.\n\nif mod is 0: steps is max_triples.\n\nif mod is 1:\n\n   if max_triples is zero, then we need to handle it. Because if f is 1, it's invalid. But if f is 4 (like 3 +1), but we can't have 1. So here, we need to take one less tri",
    "### Question:\nYou are given a 0-indexed integer array nums of length n where n is the total number of students in the class. The class teacher tries to select a group of students so that all the students remain happy.\nThe i^th student will become happy if one of these two conditions is met:\n\nThe student is selected and the total number of selected students is strictly greater than nums[i].\nThe student is not selected and the total number of selected students is strictly less than nums[i].\n\nReturn the number of ways to select a group of students so that everyone remains happy.\n \nExample 1:\n\nInput: nums = [1,1]\nOutput: 2\nExplanation: \nThe two possible ways are:\nThe class teacher selects no student.\nThe class teacher selects both students to form the group. \nIf the class teacher selects just one student to form a group then the both students will not be happy. Therefore, there are only two possible ways.\n\nExample 2:\n\nInput: nums = [6,0,3,3,6,7,2,7]\nOutput: 3\nExplanation: \nThe three possible ways are:\nThe class teacher selects the student with index = 1 to form the group.\nThe class teacher selects the students with index = 1, 2, 3, 6 to form the group.\nThe class teacher selects all the students to form the group.\n\n \nConstraints:\n\n1 <= nums.length <= 10^5\n0 <= nums[i] < nums.length\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def countWays(self, nums: List[int]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the number of ways to select a group of students such that all are happy based on the given conditions. Let me try to understand the problem first.\n\nSo each student has two conditions based on whether they are selected or not. Let me rephrase:\n\nFor the i-th student:\n- If selected, then the total selected count must be > nums[i].\n- If not selected, then the total selected count must be < nums[i].\n\nThe goal is to count all possible group sizes (let's call it k) such that when exactly k students are selected, all students are happy. Then, the answer is the number of valid k's.\n\nWait, but how do I ensure that for a particular k (number of selected students), all the students meet their respective conditions?\n\nLet me think. For a given k, for each student in the array:\n\nIf the student is selected: then k must be > nums[i].\n\nIf the student is not selected: then k must be < nums[i].\n\nBut how do I know for each student whether they are selected or not? Because when we choose a group of size k, some students are selected and some are not. But for each student, depending on whether they are in the group or not, their condition applies.\n\nBut wait, the problem isn't about choosing exactly k students. Wait, no, the group can be any size. But the problem is to find all possible group sizes k such that when exactly k students are selected, there exists at least one group of size k where all students meet their conditions.\n\nWait, no. The problem is to count the number of ways to select a group. But different groups of the same size may exist. However, the conditions here are such that for a given k, if there exists a way to select a group of size k that satisfies all students, then all possible groups of that size that meet the conditions contribute to the count. But how?\n\nWait, no. The problem says that for all students to be happy. So for a given k, every student must satisfy either:\n\nIf they are in the group (selected), then k > nums[i].\n\nOr if they are not in the group, then k < nums[i].\n\nSo for a group of size k to be valid, when you look at each student in the group, their nums[i] must be less than k, and for those not in the group, their nums[i] must be greater than k.\n\nBut how can we ensure that there exists such a group? Because the selection of which students are in the group affects this.\n\nWait, let's think. Suppose we sort the array. Let's say the array is sorted. Then, perhaps for a certain k, if we select some students such that all selected have nums[i] < k (no, wait the condition is that if selected, then k>nums[i]. So selected students must have nums[i] < k. Similarly, unselected students must have nums[i] > k. Wait no: if a student is not selected, then the number of selected students (k) must be less than nums[i]. So for that student, nums[i] must be greater than k?\n\nNo, the condition for not selected is that the total selected (k) is strictly less than nums[i]. So for an unselected student, nums[i] must be > k.\n\nWait, so for a given k, all students must fall into one of these two categories:\n\nEither they are selected (so k > nums[i]) or they are not selected (so k < nums[i]).\n\nWait, but if a student is selected, then k must be > nums[i]. And if they are not selected, then k must be < nums[i]. So for each student, nums[i] must be less than k or greater than k. Wait, but what if nums[i] equals k? Then for that student, if selected, k is not > nums[i], and if not selected, k is not < nums[i]. So that student can't be happy. Therefore, for any student with nums[i] == k, the group is invalid.\n\nTherefore, for a group size k to be valid, there must be no students with nums[i] == k. Because if there is such a student, they can't be happy regardless of whether they are selected or not.\n\nSo first, for any k where there exists a student with nums[i] = k, then no group of size k can be valid. Therefore, such k's are invalid.\n\nNow, for the other cases where there are no students with nums[i] equal to k, we need to select some group of size k such that:\n\nAll selected students have nums[i] < k.\n\nAll unselected students have nums[i] > k.\n\nWait, let's think. Suppose we sort the array. Let's say we have sorted nums. For a given k, if we can select exactly k students such that all selected ones have nums[i] < k, and all unselected ones (n - k students) have nums[i] > k. How?\n\nBecause if that's the case, then:\n\nFor selected students, since k (the count) is > their nums[i], their condition is satisfied.\n\nFor unselected students, since their nums[i] is > k, the count (k) is less than their nums[i], so their condition is satisfied.\n\nBut how do we ensure that this is possible?\n\nLet me think. Let's sort the array. Suppose the array is sorted in non-decreasing order.\n\nFor a given k, if we consider that there are exactly m students with nums[i] < k, and p students with nums[i] > k, and the rest (if any) are equal to k (but we already checked that there are none equal to k).\n\nBut since there are no students equal to k, m + p = n.\n\nIn this case, to select a group of size k students, we need to select exactly k students. But all selected students must have nums[i] < k, so we need to select k students from those that have nums[i] < k. However, how many are there?\n\nSuppose there are x students with nums[i] < k. Then, in order to select k students, we need x >= k. Because we have to select k students all from that set. Also, the unselected students are (n - k) students, which must all be in the set with nums[i] > k. How many students have nums[i] > k? Since total is n, and there are x students with nums[i] < k, then the number of students with nums[i] > k is n - x - (number of students with nums[i] =k). But since we already checked that there are zero students with nums[i] =k, then it's n -x. So for the unselected students, the number of students with nums[i] >k is n -x. Since the unselected students are (n -k), we need that n -x >= (n -k). Which simplifies to x >=k.\n\nWait, because:\n\nNumber of students with nums[i] >k is (n -x). And the number of unselected students is (n -k). For all of them to have nums[i] >k, the unselected students must be exactly all of the students with nums[i] >k. Which requires that (n -k) <= (n -x). Which simplifies to x >=k.\n\nSo combining both conditions: to select k students (all with nums[i] <k) and leave the rest (with nums[i] >k), we need that the number of students with nums[i] <k is >=k, and the number of students with nums[i] >k is >= (n -k). But since the number of students with nums[i] >k is (n -x) where x is the count of nums[i] <k,",
    "### Question:\nYou are given a 0-indexed array maxHeights of n integers.\nYou are tasked with building n towers in the coordinate line. The i^th tower is built at coordinate i and has a height of heights[i].\nA configuration of towers is beautiful if the following conditions hold:\n\n1 <= heights[i] <= maxHeights[i]\nheights is a mountain array.\n\nArray heights is a mountain if there exists an index i such that:\n\nFor all 0 < j <= i, heights[j - 1] <= heights[j]\nFor all i <= k < n - 1, heights[k + 1] <= heights[k]\n\nReturn the maximum possible sum of heights of a beautiful configuration of towers.\n \nExample 1:\n\nInput: maxHeights = [5,3,4,1,1]\nOutput: 13\nExplanation: One beautiful configuration with a maximum sum is heights = [5,3,3,1,1]. This configuration is beautiful since:\n- 1 <= heights[i] <= maxHeights[i]  \n- heights is a mountain of peak i = 0.\nIt can be shown that there exists no other beautiful configuration with a sum of heights greater than 13.\nExample 2:\n\nInput: maxHeights = [6,5,3,9,2,7]\nOutput: 22\nExplanation: One beautiful configuration with a maximum sum is heights = [3,3,3,9,2,2]. This configuration is beautiful since:\n- 1 <= heights[i] <= maxHeights[i]\n- heights is a mountain of peak i = 3.\nIt can be shown that there exists no other beautiful configuration with a sum of heights greater than 22.\nExample 3:\n\nInput: maxHeights = [3,2,5,5,2,3]\nOutput: 18\nExplanation: One beautiful configuration with a maximum sum is heights = [2,2,5,5,2,2]. This configuration is beautiful since:\n- 1 <= heights[i] <= maxHeights[i]\n- heights is a mountain of peak i = 2. \nNote that, for this configuration, i = 3 can also be considered a peak.\nIt can be shown that there exists no other beautiful configuration with a sum of heights greater than 18.\n\n \nConstraints:\n\n1 <= n == maxHeights <= 10^3\n1 <= maxHeights[i] <= 10^9\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def maximumSumOfHeights(self, maxHeights: List[int]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the maximum possible sum of heights of a beautiful configuration of towers. Let's think about what the problem says.\n\nA beautiful configuration has two main conditions. First, each height must be between 1 and the corresponding maxHeight. Second, the heights array must be a mountain array. A mountain array has a peak such that to the left it's non-decreasing and to the right non-increasing.\n\nThe goal is to find the maximum sum of such a configuration. So how do I approach this?\n\nHmm. The key idea here is that for each possible peak position, I need to determine the maximum possible heights that satisfy the mountain condition with that peak. Then, among all possible peaks, choose the one that gives the maximum sum.\n\nSo for each index i (the peak), I need to construct the maximum possible heights array where the heights increase (or stay same) up to i, then decrease (or stay same) after i, without exceeding maxHeights at any position. Then calculate the sum for each such possible configuration and take the maximum.\n\nBut how do I efficiently compute this for each possible peak?\n\nLet me think. For each peak position i, the maximum valid heights would be such that:\n\n- For the left side (from 0 to i), each element is as large as possible but not larger than the previous one. Similarly for the right side (from i to n-1).\n\nBut how to compute that?\n\nAn approach could be for each position i, compute the maximum possible height for each position to the left of i (including i) such that they are non-decreasing, and each is <= their respective maxHeights. Then similarly for the right side, ensuring non-increasing. Then combine the two parts.\n\nFor example, for a given peak i, the height at i is the minimum between its maxHeight and the minimum of the maximums from the left and right directions. Wait, maybe not. Let me think again.\n\nLet me take an example. Let's say I want to build up to the peak at i. For the left part (from 0 to i), each element can be at most min( its own maxHeight, the next element to the right (since they must be non-decreasing). But how to compute this?\n\nA common approach for such problems is to compute for each position, the maximum possible value considering the left and right constraints.\n\nLet me think about two passes for each possible peak. But since the peak can be any position, perhaps we can precompute for each position, the maximum possible non-decreasing sequence up to that position (from the left), and the maximum possible non-increasing sequence from that position (to the right). Then for each possible peak, combine the left part up to it and the right part from it.\n\nBut how to do this efficiently?\n\nAlternatively, for each possible peak, construct the array as follows:\n\nLeft part (from 0 to peak):\n\nFor each position j from 0 to peak, the height[j] is the minimum between maxHeights[j] and height[j+1]. Because since the array must be non-decreasing up to the peak, each element can't be larger than the next one. So starting from the peak, which can be as high as possible (but constrained by the right part as well?), perhaps we can build the left part by making sure each previous element is at most the min of its max and the next element. Similarly for the right part.\n\nWait, here's an idea. For a given peak position i, the maximum possible heights for the left part (from 0 to i) would be such that each height is as high as possible, but not higher than the next one (since they need to be non-decreasing), and not higher than their maxHeights. So starting from the peak, which can be at most maxHeights[i], then for j = i-1 downto 0, each height[j] is the minimum between maxHeights[j] and the previous height (height[j+1]). Because that way, the sequence is non-decreasing and each is as high as possible.\n\nSimilarly, for the right part (from i to end), for each position j from i to n-1, the height[j] is the minimum between maxHeights[j] and the previous height (height[j-1]). This ensures non-increasing sequence.\n\nSo combining these two parts for each possible peak, calculate the sum, and track the maximum.\n\nYes! That makes sense. For example, in the first sample input:\n\nmaxHeights = [5,3,4,1,1], peak at 0.\n\nLeft part is just [5], since it's the peak. Then the right part from 0 to end:\n\nposition 0 is 5. Position 1 can be min(3, 5) → 3. Position 2: min(4,3) → 3. Position 3: min(1,3) →1. Position4: min(1,1) →1. The total sum is 5+3+3+1+1=13, which matches the example.\n\nSo for each possible peak, we need to compute this left and right part and sum them, then take the maximum.\n\nNow the problem is, how to implement this efficiently.\n\nThe straightforward approach would be for each possible peak in 0..n-1, compute the left part (non-decreasing up to peak) and right part (non-increasing after peak), then sum all. Take the maximum over all peaks.\n\nBut how to compute for each peak?\n\nThe steps for each peak i would be:\n\n- For the left part (from 0 to i):\n\n   heights_left[0] = min(maxHeights[0], ... ?\n\nWait, for the left part up to i, the maximum possible non-decreasing sequence. How to compute it? For each j from 0 to i, heights_left[j] can be at most maxHeights[j], and also can't exceed heights_left[j+1], since when moving left from the peak, each previous element can't be higher than the next one.\n\nSo for the left part, starting from the peak, moving left to 0:\n\nheights_left[i] = maxHeights[i]\n\nfor j from i-1 downto 0:\n\n   heights_left[j] = min( maxHeights[j], heights_left[j+1] )\n\nSimilarly, for the right part, starting from i, moving right to n-1:\n\nheights_right[i] = maxHeights[i]\n\nfor j from i+1 to n-1:\n\n   heights_right[j] = min( maxHeights[j], heights_right[j-1] )\n\nThen combine the two parts (taking left part up to i, and right part from i onwards) and sum all elements.\n\nBut for each possible peak i, we need to compute this. Then take the maximum.\n\nThe time complexity would be O(n^2), since for each of n peaks, we process O(n) elements. Since n can be up to 1e3, O(1e6) operations would be accept",
    "### Question:\nYou have n processors each having 4 cores and n * 4 tasks that need to be executed such that each core should perform only one task.\nGiven a 0-indexed integer array processorTime representing the time at which each processor becomes available for the first time and a 0-indexed integer array tasks representing the time it takes to execute each task, return the minimum time when all of the tasks have been executed by the processors.\nNote: Each core executes the task independently of the others.\n \nExample 1:\n\nInput: processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]\nOutput: 16\nExplanation: \nIt's optimal to assign the tasks at indexes 4, 5, 6, 7 to the first processor which becomes available at time = 8, and the tasks at indexes 0, 1, 2, 3 to the second processor which becomes available at time = 10. \nTime taken by the first processor to finish execution of all tasks = max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16.\nTime taken by the second processor to finish execution of all tasks = max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13.\nHence, it can be shown that the minimum time taken to execute all the tasks is 16.\nExample 2:\n\nInput: processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]\nOutput: 23\nExplanation: \nIt's optimal to assign the tasks at indexes 1, 4, 5, 6 to the first processor which becomes available at time = 10, and the tasks at indexes 0, 2, 3, 7 to the second processor which becomes available at time = 20.\nTime taken by the first processor to finish execution of all tasks = max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.\nTime taken by the second processor to finish execution of all tasks = max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.\nHence, it can be shown that the minimum time taken to execute all the tasks is 23.\n\n \nConstraints:\n\n1 <= n == processorTime.length <= 25000\n1 <= tasks.length <= 10^5\n0 <= processorTime[i] <= 10^9\n1 <= tasks[i] <= 10^9\ntasks.length == 4 * n\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def minProcessingTime(self, processorTime: List[int], tasks: List[int]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. The problem is about assigning tasks to processors' cores such that each core does exactly one task, and we need to find the minimum time when all tasks are completed. Hmm.\n\nSo, each processor has 4 cores, and there are n processors. The total number of tasks is 4n, which matches exactly. So for each processor, we need to assign 4 tasks, right?\n\nThe goal is to arrange the tasks so that the maximum (over all cores) of (processor available time + task time) is as small as possible.\n\nWait, how to approach this? Let me think about the example given. In example 1, the processorTime is [8,10], and tasks are sorted in a way that the largest tasks are assigned to the earliest available processor. Oh right, because if you sort the tasks in descending order, and then assign the first 4 to the first processor (which becomes available earliest), then the next 4 to the next processor, etc. But wait, how does that work?\n\nWait, let's think about the example 1. The tasks are sorted as [8,7,5,4,3,2,2,1]. Then processor 0 (available at 8) takes 8,7,5,4. Their completion times are 8+8=16, 8+7=15, 8+5=13, 8+4=12. The max here is 16. The second processor (available at 10) takes 3,2,2,1. Their times are 10+3=13, etc. So overall max is 16. \n\nIf we instead assigned smaller tasks to the earlier processor, maybe the max would be higher. Like, if processor 0 takes 1,2,3,5. Then the max for that processor would be 8+5=13, but then the other processor would have bigger tasks. Wait no, because the tasks are split in a way that the largest ones are handled by the earliest available processors. \n\nSo the strategy seems to be: sort the processorTime in ascending order (so earliest available first), and sort the tasks in descending order. Then assign the first 4 tasks (largest) to the first processor, next 4 to the next, etc. Then for each processor, the available time plus the task's times, the max among those four. Then the overall max among all processors.\n\nYes, because for each processor, the largest task assigned to it will determine its completion time. So by pairing the largest tasks with the earliest available processors, we can minimize the maximum.\n\nSo the steps would be:\n\n1. Sort the processorTime in ascending order. Because we want to assign the earliest available processors the largest tasks, so that their processing time plus the task's time is minimized.\n\nWait, no. Let me think again. Suppose a processor is available earlier. If we assign larger tasks to it, since it's available earlier, the total time would be (earlier available time + larger task time) versus if we assign smaller tasks to it (earlier + smaller). But perhaps another processor which is available later but has smaller tasks may result in a lower max. \n\nBut the idea is that for each processor, the maximum task assigned to it will determine its finish time. So to minimize the overall maximum, we need to balance. \n\nThis seems like a classic problem where you have to match the largest tasks to the earliest available resources. Let me think of an example. \n\nSuppose two processors: one available at time 0, another at time 10. Suppose tasks are 10, 9, 1, 1. If we assign 10 and 9 to the first processor (available at 0), then their finish times are 0+10=10 and 0+9=9. Max is 10. The second processor has 1 and 1. Its max is 10+1=11. Overall max is 11. But if we assign 10 and 1 to the first processor, and 9 and 1 to the second. Then first processor's max is 0+10=10. Second's is 10+9=19. Which is worse. Or if we assign 1 and 1 to the first, and 10 and 9 to the second. Then first's max is 0+1=1, second's 10+10=20. Which is worse. So the best approach was to assign the two largest tasks to the earliest processor. \n\nSo the approach is correct. Therefore, the algorithm should be:\n\n- Sort the processors' available times in increasing order. \n- Sort the tasks in descending order. \n- For each processor i, take the 4*i to 4*i +3 tasks (the next 4 largest tasks). Compute the task time plus the processor's available time. Keep track of the maximum among all these.\n\nWait, but for each processor, the maximum of the four tasks assigned to it (since each core has one task) is the available time plus the maximum of the four task times. Because the four tasks are added to the processor's available time, and the max of those four will be the available time plus the maximum task time among them. \n\nSo, for example, if a processor is available at time T, and we assign four tasks with times t0, t1, t2, t3. The max is T + max(t0, t1, t2, t3). So for each group of four tasks assigned to a processor, we need to compute T_i + max of the group.\n\nTherefore, the problem reduces to grouping the tasks into groups of four, and assigning each group to a processor in such a way that when we compute the (processor's available time + group's max), the overall maximum across all groups is minimized.\n\nSo how to do that?\n\nSince the processors are sorted in increasing order (earliest available first), and the tasks are sorted in descending order, then group the first four as the first group (largest four tasks), assign to first processor (earliest available). Then next four (next largest) to second processor, etc. \n\nBecause for each group, the largest task in the group is the first one (since the group is sorted in descending order). So for each group, the max is the first element of the group. \n\nSo for example, after grouping, for the first group (tasks 0-3, sorted descending), max is first element. \n\nWait, let's think. Suppose the tasks are sorted in descending order. Then in a group of four, the first element is the maximum. So when we assign a group to a processor, the finish time is (processor available time + task[0] of the group). \n\nThus, the algorithm steps would be:\n\nSort processorTime in ascending order.\n\nSort tasks in descending order.\n\nFor each i from 0 to n-1:\n\n   group i (tasks[4*i ... 4*i +3]) is assigned to processor i.\n\n   The finish time for thi",
    "### Question:\nYou are given a 0-indexed integer array nums and a positive integer k.\nYou can do the following operation on the array any number of times:\n\nChoose any two distinct indices i and j and simultaneously update the values of nums[i] to (nums[i] AND nums[j]) and nums[j] to (nums[i] OR nums[j]). Here, OR denotes the bitwise OR operation, and AND denotes the bitwise AND operation.\n\nYou have to choose k elements from the final array and calculate the sum of their squares.\nReturn the maximum sum of squares you can achieve.\nSince the answer can be very large, return it modulo 10^9 + 7.\n \nExample 1:\n\nInput: nums = [2,6,5,8], k = 2\nOutput: 261\nExplanation: We can do the following operations on the array:\n- Choose i = 0 and j = 3, then change nums[0] to (2 AND 8) = 0 and nums[3] to (2 OR 8) = 10. The resulting array is nums = [0,6,5,10].\n- Choose i = 2 and j = 3, then change nums[2] to (5 AND 10) = 0 and nums[3] to (5 OR 10) = 15. The resulting array is nums = [0,6,0,15].\nWe can choose the elements 15 and 6 from the final array. The sum of squares is 15^2 + 6^2 = 261.\nIt can be shown that this is the maximum value we can get.\n\nExample 2:\n\nInput: nums = [4,5,4,7], k = 3\nOutput: 90\nExplanation: We do not need to apply any operations.\nWe can choose the elements 7, 5, and 4 with a sum of squares: 7^2 + 5^2 + 4^2 = 90.\nIt can be shown that this is the maximum value we can get.\n\n \nConstraints:\n\n1 <= k <= nums.length <= 10^5\n1 <= nums[i] <= 10^9\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def maxSum(self, nums: List[int], k: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's try to figure out this problem. So, the task is to perform some operations on the array and then select k elements to maximize the sum of their squares. The operations allowed are choosing two indices i and j, and setting nums[i] to (nums[i] AND nums[j]) and nums[j] to (nums[i] OR nums[j]). Hmm, interesting. \n\nFirst, I need to understand what these operations do. Let me think about the effect of such an operation. Let's say I have two numbers a and b. After the operation, the new values are (a & b) and (a | b). Wait, but what's the sum of these two? Let me check. Because (a & b) + (a | b) equals a + b. Because a & b is the bits where both are 1, and a | b is the bits where at least one is 1. Adding them together would be (a & b) + (a | b) = (a & b) + (a | b) = a + b. Because for each bit position, if both are 1, then in the sum it's 2*bit. Otherwise, it's 1*bit. But adding a and b directly gives the same result. Oh right! So the sum of the two numbers remains the same after the operation. So this operation doesn't change the sum of the array. But what about individual bits?\n\nThe key here is that the operation allows us to redistribute the bits between positions. For example, if a has a 1 in a certain bit and b has 0, then after the operation, maybe we can move that bit around. But how?\n\nLet me think about the bitwise properties. The operation replaces a and b with (a & b) and (a | b). But since a | b is equal to a + b - a & b, perhaps there's some way to manipulate bits here. However, the important thing is that the sum of all elements remains the same. Because the sum of the two elements after the operation is the same as before. Since each operation preserves the sum, the total sum of the array is fixed. However, the problem isn't about the sum, but about maximizing the sum of squares of the top k elements.\n\nAh right. The sum of squares is maximized when the selected elements are as large as possible. Because squares grow rapidly with larger numbers. So the problem reduces to redistributing the bits in such a way that we can create as many large numbers as possible, then pick the top k.\n\nSo the challenge is to figure out what's the optimal way to arrange the bits. How can the operations affect the bits in the array?\n\nLet me think about the bits for each position. Let's consider each bit position independently. Suppose we have m numbers that have a 1 in the 2^b position. Since the operations can redistribute bits, perhaps we can arrange for the maximum possible numbers to have as many high bits set as possible.\n\nWait, but how? Let me think. For example, if I have multiple bits in different positions, how can the operations help combine them?\n\nLet me take two numbers. Suppose a has a bit set in position i, and b has a bit set in position j. If I perform the operation, then after the operation, maybe we can collect bits into some numbers. For example, if I have two numbers, say 5 (101) and 3 (011). Performing the operation on them would give (101 AND 011) = 001, and (101 OR 011) = 111. So the first becomes 1 and the second becomes 7. Now, the 7 has more bits set. So through operations, we can take numbers and create numbers with higher bits by combining their bits, but perhaps losing some lower bits in the process.\n\nBut the key idea here is that for each bit position, the total count of 1s across the array is preserved. Because when you perform the operation, the sum of all the bits in a particular position across all elements remains the same. Let me check that. Suppose we look at a particular bit position. Let's say that in the original a and b, this bit is x and y respectively. After performing the operation, what is the total number of 1s in this bit?\n\nThe new values are (x AND y) and (x OR y). For this bit, the total before is x + y. After the operation, it's (x & y) + (x | y) = x + y. Because for any x and y, (x & y) + (x | y) = x + y. Therefore, for each bit position, the total count of 1s in that position across the array is fixed. So no matter how we perform operations, the total number of 1s in each bit position is fixed.\n\nThis seems like a crucial insight. For example, if there are c_b bits set in the b-th position across all numbers, then regardless of operations, this remains the same. \n\nNow, how can we use this to maximize the sum of squares? Well, since we want to maximize the sum of squares of k elements, we want those elements to have as much value as possible. To maximize the sum of squares, we should maximize each of the top k elements. How can we arrange the bits to make the top k elements as large as possible?\n\nSince each bit position's count is fixed, the maximum possible values for the elements would be achieved by setting, for each bit position, as many elements as possible to have that bit set. But how?\n\nFor example, suppose for each bit position, we can have up to m_b 1s. Let's imagine that we want to create numbers that have as many high bits as possible. For each bit from highest to lowest, we can assign the bits to the first m_b numbers. For example, if the highest bit has c_ highest count of m, then we can set those m numbers to have that bit. Then for the next bit, we set as many as possible (up to its count) to the first m numbers (so that they can have both bits set), and so on.\n\nThis sounds like constructing the maximum possible numbers by greedily assigning bits. For example, if for each bit position, we have a count of how many times it can be set. Then, for each of the top k elements, we want to assign as many bits as possible.\n\nTherefore, the maximum possible elements would be constructed by, for each bit position, contributing as much as possible to the top elements. For example, if for bit position b, there are c_b 1s available. Then, the first min(c_b, k) elements can have this bit set. Then, the value of each of those elements would include 2^b.\n\nBut how do we compute the maximum possible sum of squares? Let's think. The maximum sum would be achieved when each of the top k elements is as large as possible. So, for each bit position, we can contribute that bit to as many of the top elements as possible. \n\nLet me think with an example. Suppose for each bit position, we have counts. For example, in Example 1, the original array is [2,6,5,8]. Let's look at each bit:\n\nLet's write each number in binary:\n\n2: 10\n6: 110\n5: 101\n8: 1000\n\nLooking at each bit position:\n\nLet's list the cou",
    "### Question:\nYou are given positive integers n and m.\nDefine two integers, num1 and num2, as follows:\n\nnum1: The sum of all integers in the range [1, n] that are not divisible by m.\nnum2: The sum of all integers in the range [1, n] that are divisible by m.\n\nReturn the integer num1 - num2.\n \nExample 1:\n\nInput: n = 10, m = 3\nOutput: 19\nExplanation: In the given example:\n- Integers in the range [1, 10] that are not divisible by 3 are [1,2,4,5,7,8,10], num1 is the sum of those integers = 37.\n- Integers in the range [1, 10] that are divisible by 3 are [3,6,9], num2 is the sum of those integers = 18.\nWe return 37 - 18 = 19 as the answer.\n\nExample 2:\n\nInput: n = 5, m = 6\nOutput: 15\nExplanation: In the given example:\n- Integers in the range [1, 5] that are not divisible by 6 are [1,2,3,4,5], num1 is the sum of those integers = 15.\n- Integers in the range [1, 5] that are divisible by 6 are [], num2 is the sum of those integers = 0.\nWe return 15 - 0 = 15 as the answer.\n\nExample 3:\n\nInput: n = 5, m = 1\nOutput: -15\nExplanation: In the given example:\n- Integers in the range [1, 5] that are not divisible by 1 are [], num1 is the sum of those integers = 0.\n- Integers in the range [1, 5] that are divisible by 1 are [1,2,3,4,5], num2 is the sum of those integers = 15.\nWe return 0 - 15 = -15 as the answer.\n\n \nConstraints:\n\n1 <= n, m <= 1000\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def differenceOfSums(self, n: int, m: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to find the difference between the sum of numbers not divisible by m and the sum of those that are divisible by m. Hmm. So the problem is to calculate num1 - num2 where num1 is the sum of all numbers from 1 to n not divisible by m, and num2 is the sum of those divisible by m.\n\nWait, right. So first, I need to separate the numbers into two groups: those divisible by m and those not. Then sum each group and subtract the second sum from the first.\n\nBut how to calculate these sums efficiently without iterating through each number? Because for large n and m, like up to 1000, it's manageable even with a loop. But maybe there's a mathematical way to do it.\n\nLet me think. The total sum of numbers from 1 to n is known. It's the formula (n*(n+1))/2. Let's call that total_sum. Then, the sum of numbers divisible by m is num2. Once we have num2, num1 is total_sum - num2. Then the result would be (total_sum - num2) - num2 = total_sum - 2*num2. Oh right! So instead of calculating both sums separately, I can just compute the total sum once and then find num2. Then the answer is total_sum minus twice num2. That's clever and might save some computation time.\n\nSo the problem reduces to calculating the sum of multiples of m up to n. How to calculate that?\n\nWell, the numbers divisible by m up to n are m, 2m, 3m, ... up to km where km ≤ n. Let's find k, which is the maximum integer such that km ≤n. So k is floor(n/m). For example, if n=10 and m=3, then k is 3 (since 3*3=9 ≤10). The sum of these numbers is m*(1 + 2 + ... +k) = m * k*(k+1)/2. So that's the formula for num2.\n\nSo putting it all together:\n\ntotal_sum = n*(n+1)//2\nk = n // m\nnum2 = m * k*(k+1) // 2\nresult = (total_sum - num2) - num2\nreturn result\n\nWhich simplifies to total_sum - 2*num2.\n\nYes, that's efficient. Let me check with the examples.\n\nFirst example: n=10, m=3. total_sum is 10*11/2 = 55. k is 3. num2 is 3*(3*4)/2 = 3*6=18. Then result is 55 - 2*18 = 55-36=19. Which matches the example.\n\nSecond example: n=5, m=6. total_sum is 15. k is 0. So num2 is 0. Result is 15 - 0 = 15. Correct.\n\nThird example: n=5, m=1. k is 5. num2 is 1*5*6/2 = 15. total_sum is 15. So result is 15 - 2*15 = 15-30= -15. Which matches.\n\nGreat. So this approach works. So the steps are:\n\nCalculate the total sum of 1 to n.\n\nCalculate how many multiples of m are there up to n (k = n//m). \n\nCalculate the sum of those multiples (num2) as m * k*(k+1)/2.\n\nSubtract twice that sum from the total sum.\n\nSo the code can be written using these formulas.\n\nNow, implementing this in Python. Let's code.\n\nThe function differenceOfSums takes n and m.\n\nFirst compute total_sum = n*(n+1) // 2.\n\nThen compute k = n // m.\n\nnum2 = m * k * (k + 1) // 2.\n\nresult = total_sum - 2 * num2.\n\nReturn result.\n\nBut wait, what if m is 0? Well, the constraints say m is a positive integer, so no problem.\n\nWhat if m is larger than n? Then k is 0, num2 is 0. So result is total_sum. Which is correct.\n\nWhat if m is 1? Then all numbers are divisible by 1, so num2 is total_sum. So result is total_sum - 2 * total_sum = -total_sum. Which is correct.\n\nSo this covers all cases.\n\nSo the code would be:\n\nclass Solution:\n    def differenceOfSums(self, n: int, m: int) -> int:\n        total_sum = n * (n + 1) // 2\n        k = n // m\n        num2 = m * k * (k + 1) // 2\n        return total_sum - 2 * num2\n\nYes. Let's test with the examples again.\n\nExample 3: n=5, m=1. total_sum is 15. k=5. num2 = 1*5*6//2 = 15. 15 - 2*15 = -15. Correct.\n\nYes. Looks good.\n\nThis approach is O(1) time and space, which is efficient.</think>To solve this problem, we need to determine the difference between the sum of integers in the range [1, n] that are not divisible by a given integer m and the sum of those that are divisible by m. \n\n### Approach\n1. **Total Sum Calculation**: First, calculate the total sum of all integers from 1 to n using the formula for the sum of the first n natural numbers, which is $ \\text{total\\_sum} = \\frac{n(n+1)}{2} $.\n\n2. **Identify Multiples of m**: Determine how many integers up to n are multiples of m. Let $ k $ be the count of these multiples, which can be found as $ k = \\left\\lfloor \\frac{n}{m} \\right\\rfloor $.\n\n3. **Sum of Multiples of m**: Calculate the sum of these multiples. The sum of the first $ k $ multiples of m is given by $ \\text{num2} = m \\times \\frac{k(k+1)}{2} $.\n\n4. **Compute the Result**: The result is derived by subtracting twice the sum of the multiples of m (num2) from the total sum. This is because the sum of non-multiples (num1) can be expressed as $ \\text{total\\_sum} - \\text{num2} $, and the desired result is $ \\text{num1} - \\text{num2} $, which simplifies to $ \\text{total\\_sum} - 2 \\times \\text{num2} $.\n\n### Solution Code\n```python\nclass Solution:\n    def differenceOfSums(self, n: int, m: int) -> int:\n        total_sum = n * (n + 1) // 2\n        k = n // m\n        num2 = m * k * (k + 1) // 2\n        return total_sum - 2 * num2\n```\n\n### Explanation\n1. **Total Sum Calculation**: The formula $ \\frac{n(n+1)}{2} $ efficiently computes the sum of the first n natural numbers.\n2. **Multiples Identification**: By dividing n by m using integer division, we determine how many multiples of m exist up to n.\n3. **Sum of Multiples**: Using the arithmetic series sum formula, we adjust for the multiples of m by incorporating m into the calculation.\n4. **Result Calculation**: By leveraging the derived formula, we efficiently compute the required difference without explicitly iterating through the range, resulting in an O(1) time complexity solution. This approach ensures optimal performance even for large values of n and m within the given constraints.",
    "### Question:\nThere are n teams numbered from 0 to n - 1 in a tournament.\nGiven a 0-indexed 2D boolean matrix grid of size n * n. For all i, j that 0 <= i, j <= n - 1 and i != j team i is stronger than team j if grid[i][j] == 1, otherwise, team j is stronger than team i.\nTeam a will be the champion of the tournament if there is no team b that is stronger than team a.\nReturn the team that will be the champion of the tournament.\n \nExample 1:\n\nInput: grid = [[0,1],[0,0]]\nOutput: 0\nExplanation: There are two teams in this tournament.\ngrid[0][1] == 1 means that team 0 is stronger than team 1. So team 0 will be the champion.\n\nExample 2:\n\nInput: grid = [[0,0,1],[1,0,1],[0,0,0]]\nOutput: 1\nExplanation: There are three teams in this tournament.\ngrid[1][0] == 1 means that team 1 is stronger than team 0.\ngrid[1][2] == 1 means that team 1 is stronger than team 2.\nSo team 1 will be the champion.\n\n \nConstraints:\n\nn == grid.length\nn == grid[i].length\n2 <= n <= 100\ngrid[i][j] is either 0 or 1.\nFor all i grid[i][i] is 0.\nFor all i, j that i != j, grid[i][j] != grid[j][i].\nThe input is generated such that if team a is stronger than team b and team b is stronger than team c, then team a is stronger than team c.\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def findChampion(self, grid: List[List[int]]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. The problem is to find the champion team in a tournament based on a given grid. The champion is a team that no other team is stronger than them. Hmm. So the grid is such that grid[i][j] is 1 if team i is stronger than j, otherwise 0. And since the input ensures that the relations are transitive (like if a is stronger than b and b than c, then a is stronger than c), there must be a unique champion. Wait, but how do I find it?\n\nWait, maybe there's a way to determine the champion without checking all possibilities. Let me think. For each team, if they have all 1's in their row (except their own position, which is 0), then they are stronger than all other teams. But that's probably not the case for all teams except the champion. Wait, but there must be exactly one team that is the champion. How can I find it?\n\nAnother approach: The champion team must have all 1's in their row except for their own entry. Because if they are stronger than every other team. So for each row i, check if all elements are 1 except grid[i][i]. But how do I find which row that is?\n\nAlternatively, since for any two teams a and b, exactly one of grid[a][b] or grid[b][a] is 1. And because the relations are transitive, there must be one team that beats all others. So how do I find that team?\n\nSo the algorithm could be: For each team, count how many teams it beats. The champion would have (n-1) count, since they beat all others. So the team whose row has sum (n-1) is the champion. Because each entry in the row is 1 for every team they are stronger than.\n\nYes! Because the sum of the row for team i is the number of teams they are stronger than. Since the champion is stronger than all others, their row sum would be n-1.\n\nSo the approach is: iterate through each row, calculate the sum of the elements. The row with sum equal to (n-1) is the champion.\n\nWait, let me check with the examples.\n\nFirst example: grid is [[0,1],[0,0]]. n is 2. So for row 0, sum is 1 (which is 2-1=1). Row 1's sum is 0. So the answer is 0. Which matches the example.\n\nSecond example: grid is [[0,0,1],[1,0,1],[0,0,0]]. n is 3. The rows sums are:\n\nRow 0: 0 + 0 + 1 = 1. Not 2.\n\nRow 1: 1 + 0 + 1 = 2. Which is 3-1=2. So team 1 is the champion. Correct.\n\nRow 2: 0+0+0=0.\n\nYes. So the approach works.\n\nSo the plan is:\n\n- For each team i (each row index i), calculate the sum of grid[i].\n\n- The team with sum equal to (n-1) is the champion.\n\nSo how to implement this in Python.\n\nThe steps in code:\n\n- Get n from len(grid).\n\n- Iterate through each row (for i in range(n)), compute sum of the row.\n\n- If sum equals n-1, return that i.\n\nBut wait, since the problem says that there is exactly one champion, we can return as soon as we find it.\n\nSo code outline:\n\nclass Solution:\n\n    def findChampion(self, grid: List[List[int]]) -> int:\n\n        n = len(grid)\n\n        for i in range(n):\n\n            if sum(grid[i]) == n - 1:\n\n                return i\n\n        # But according to problem statement, there is exactly one, so this loop will always find it.\n\nYes. That's straightforward.\n\nWait, but let's check for the second example. The row for team 1 is [1,0,1]. Sum is 2, which is indeed 3-1. So correct.\n\nAnother test case: if n is 3, and the champion is team 2, then their row would have two 1's. Like [[0,0,0], [0,0,0], [1,1,0]]. Wait but according to the problem's conditions, if team 2 is stronger than team 0 and 1, then in their row, grid[2][0] is 1, grid[2][1] is 1. So sum is 2, which equals 3-1. Yes.\n\nBut how does the problem's conditions ensure that there is exactly one such team? Because the problem says that the input is generated such that if a is stronger than b and b is stronger than c, then a is stronger than c. So the relations are transitive. Therefore, there must be a unique champion. So our code will find exactly one team with sum n-1.\n\nSo this approach works.\n\nSo the code is straightforward. Let's implement it.\n\nNow, in Python, for each row in grid, sum the elements. For example, sum(grid[i]) gives the sum of the ith row.\n\nSo the code as per the starter code would be:\n\nclass Solution:\n\n    def findChampion(self, grid: List[List[int]]) -> int:\n\n        n = len(grid)\n\n        for i in range(n):\n\n            if sum(grid[i]) == n-1:\n\n                return i\n\n        return -1  # though this line should never be reached per problem constraints.\n\nBut according to the problem constraints, there is exactly one champion, so the loop will always return.\n\nTesting with the examples:\n\nExample 1: n is 2. For i=0, sum([0,1])=1=1. So return 0. Correct.\n\nExample 2: for i=1, sum([1,0,1])=2=3-1. So return 1. Correct.\n\nAnother test case: let's say grid is [[0,1,1], [0,0,1], [0,0,0]]. n=3. So team 0's sum is 2, so returns 0. Which would be correct.\n\nYes.\n\nThis seems to work. So that's the solution.</think>To solve this problem, we need to determine the champion team in a tournament based on a given grid where each entry indicates the relative strength of teams. The champion is the team that is stronger than all other teams.\n\n### Approach\nThe key insight is that the champion team will have a row in the grid where all entries (except their own) are 1, indicating they are stronger than every other team. This means the sum of the entries in the champion team's row will be exactly `n-1` (since there are `n` teams and the diagonal entry is always 0). \n\nThe steps to find the champion are:\n1. Iterate through each team's row in the grid.\n2. Calculate the sum of each row. The row with a sum of `n-1` corresponds to the champion team.\n3. Return the index of the team with this sum.\n\n### Solution Code\n```python\nclass Solution:\n    def findChampion(self, grid: List[List[int]]) -> int:\n        n = len(grid)\n        for i in range(n):\n            if sum(grid[i]) == n - 1:\n                return i\n        return -1  # According to constraints, this line should not be reached\n```\n\n### Explanation\n- **Iterate through each row**: Each row represents a team's strength relative to other teams.\n- **Sum calculation**: The sum of the entries in a row indicates how many teams the corresponding team is stronger than. A sum of `n-1` means the team is stronger than all other teams.\n- **Return the champion**: The first team (row) found with the sum of `n-1` is immediately returned as the champion, leveraging the problem's constraint that ensures a unique champion exists.\n\nThis approach efficiently identifies the champion in O(n^2) time complexity in the worst case, which is acceptable given the constraints (n ≤ 100).",
    "### Question:\nGiven three integers a, b, and n, return the maximum value of (a XOR x) * (b XOR x) where 0 <= x < 2^n.\nSince the answer may be too large, return it modulo 10^9 + 7.\nNote that XOR is the bitwise XOR operation.\n \nExample 1:\n\nInput: a = 12, b = 5, n = 4\nOutput: 98\nExplanation: For x = 2, (a XOR x) = 14 and (b XOR x) = 7. Hence, (a XOR x) * (b XOR x) = 98. \nIt can be shown that 98 is the maximum value of (a XOR x) * (b XOR x) for all 0 <= x < 2^n.\n\nExample 2:\n\nInput: a = 6, b = 7 , n = 5\nOutput: 930\nExplanation: For x = 25, (a XOR x) = 31 and (b XOR x) = 30. Hence, (a XOR x) * (b XOR x) = 930.\nIt can be shown that 930 is the maximum value of (a XOR x) * (b XOR x) for all 0 <= x < 2^n.\nExample 3:\n\nInput: a = 1, b = 6, n = 3\nOutput: 12\nExplanation: For x = 5, (a XOR x) = 4 and (b XOR x) = 3. Hence, (a XOR x) * (b XOR x) = 12.\nIt can be shown that 12 is the maximum value of (a XOR x) * (b XOR x) for all 0 <= x < 2^n.\n\n \nConstraints:\n\n0 <= a, b < 2^50\n0 <= n <= 50\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def maximumXorProduct(self, a: int, b: int, n: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to find the maximum value of (a XOR x) multiplied by (b XOR x) where x is between 0 and 2^n - 1. And then return it modulo 1e9+7. Hmm. The constraints are big here, especially since a and b can be up to 2^50, and n can be up to 50. So brute-forcing all possible x values isn't feasible for larger n, like when n is 50, there are 2^50 possibilities. That's way too big. So I need a smarter approach.\n\nFirst, let's think about the problem. For each bit in x, how does flipping it affect the product? Since XOR flips the bits where x has a 1. The idea is to find for each bit position, whether setting it in x will lead to a higher product.\n\nBut how do I decide which bits to set in x? Let's consider that a and b could be equal or different. Also, when we XOR with x, both a and b are affected. For example, flipping a certain bit in x will flip the same bit in both a^x and b^x. Wait, no. Wait, x's bits are applied once each. So, if x has a bit set, then a^x will have that bit flipped compared to a. Similarly for b^x.\n\nSo, for each bit from highest to lowest, we can decide whether to set that bit in x. Because higher bits have more impact on the value. The key is to find for each bit, whether flipping it (by setting x's bit to 1) would result in a larger product.\n\nLet me think about this step by step. Let's suppose we are building x bit by bit, starting from the highest bit to the lowest. For each bit position i, we decide whether to set it in x. The decision depends on whether it increases the product (a^x) * (b^x).\n\nLet me think about how to approach this. Let's consider that initially, x is 0. Then, for each bit position from high to low, we can try to set it in x and check if it's beneficial.\n\nBut how do we evaluate whether setting a bit in x is beneficial? Let's take an example. Suppose we have a bit at position k. If we set this bit in x, then a^x will have that bit flipped, and same for b^x.\n\nSo, let's denote current a_xor as A and current b_xor as B. Suppose we are considering setting the k-th bit in x. Then, after setting, the new values would be A ^ (1 << k) and B ^ (1 << k). The product becomes (A ^ (1<<k)) * (B ^ (1<<k)). We need to compare this with the previous product (A*B). If the new product is larger, then we should set that bit in x.\n\nBut how to do this efficiently?\n\nThe problem is that for each bit, we have to decide whether to flip it. But since bits higher than the current position have already been decided, we can process them in order.\n\nSo the plan is:\n\n- Start with x=0. So A = a, B = b.\n- For each bit from (n-1) down to 0 (since x can be up to 2^n, so bits are 0 to n-1? Wait, x is less than 2^n, so the maximum x is 2^n - 1. So the bits are from 0 to n-1. But for each bit, we can process from highest to lowest. Let's say the highest bit is (n-1)th bit.\n\nWait, but what if a and b have higher bits than n? For example, a and b could be up to 2^50, but n is 50. So x can have up to 50 bits. If n is less than the bits of a and b, then XORing with x can affect the higher bits.\n\nSo, for each bit position in x (from 0 to n-1), we need to decide whether to set it.\n\nSo the approach is:\n\nFor each bit position i (starting from the highest to the lowest):\n\n- Check if flipping this bit (i.e., setting x's bit here) would increase the product A*B, where A is current a^x and B is current b^x.\n\nIf yes, then set this bit in x. Otherwise, leave it as 0.\n\nBut how to compute this efficiently?\n\nLet's formalize this. Suppose current A and B. Let mask be the bits we have decided so far. But perhaps it's easier to track A and B as we go.\n\nLet me think. Let's take the current bit i. Suppose we are considering whether to set it in x. Let's assume that for higher bits than i, we have already made decisions. Now for bit i, we can decide.\n\nLet me think about what happens when we set this bit. The current A and B will have their i-th bit flipped. Let's compute the product in both cases (flip or not flip). If flipping gives a higher product, then we do it.\n\nBut how to compare?\n\nLet original product be A*B.\n\nIf we flip both A and B's i-th bit, then new product is (A ^ (1<<i)) * (B ^ (1<<i)).\n\nWe need to check if this is larger than the original product. If so, we take it, else not.\n\nBut how to compute this? Let's expand the new product.\n\nLet me denote:\n\nLet delta_A = (A ^ (1<<i)) - A. Well, since XORing with 1<<i will flip the bit. So if the current bit in A is 0, then delta is + (1<<i). If it's 1, delta is - (1<<i). Same for B.\n\nBut maybe it's easier to compute the difference between the two products. Let me think:\n\nOriginal product: AB.\n\nNew product: (A + delta_A) * (B + delta_B) = AB + A*delta_B + B*delta_A + delta_A * delta_B.\n\nSince delta_A and delta_B are either + or - 2^i. But since we are flipping the same bit in A and B, delta_A and delta_B are both + (if the original bit was 0) or both - (if original bit was 1) multiplied by (1<<i).\n\nWait, for example, suppose in A, the i-th bit is 0. Then, flipping it adds (1<<i) to A. Similarly for B. So delta_A = delta_B = d, where d is either + (1<<i) or - (1<<i), depending on whether the current bit in A and B was 0 or 1.\n\nWait, no. Because A is a^x_so_far. When considering flipping this bit (i.e., adding 1<<i to x), then A becomes a ^ (x_so_far | (1<<i)), which is (a ^ x_so_far) ^ (1<<i) = A ^ (1<<i). Similarly for B.\n\nSo delta_A = (A ^ (1<<i)) - A. Let me compute that. For example, if A has the i-th bit as 0, then XORing with 1<<i adds 2^i. If it's 1, then XORing subtracts 2^i. So delta_A is (if the bit is 0, +d, else -d), where d is 1<<i. Same for delta_B.\n\nSo delta_A = ( (A >> i) & 1 == 0 ) ? +d : -d. Similarly for delta_B.\n\nBut since delta_A and delta_B are both either +d or -d, depending on their current bits.\n\nSo the new product is AB + A*delta_B + B*delta_A + delta_A * delta_B.\n\nWhich equals AB + delta * (A + B) + delta^2, where delta is either +d or -d for both.\n\nWait, because delta_A = delta_B = delta. Because when you flip the same bit in x, both A and B are flipped in their i-th bit. So delta is same for both.\n\nSo delta is:\n\nif (A's i-th bit is 0) and (B's i-th bit is 0): then flipping adds d to both. So delta is +d.\n\nif (A's i-th bit is 1) and (B's i-th bit is 1): flipping subtracts d from both. delta is -d.\n\nif one is 0 and the other is 1? Well, that's impossible because when you flip the same bit in x, both A and B have their i-th bit flipped. So delta for A and B is same.\n\nSo the new product is AB + delta*(A + B) + delta^2.\n\nWe need to compare this to AB. So the difference is delta*(A + B) + delta^2. We need to see if this is positive. If yes, then flipping is better.\n\nSo, delta can be +d or -d. Let's compute for delta = d: the difference is d*(A + B) + d². For delta = -d: difference is -d*(A + B) + d².\n\nWait, but how do we know what delta is? Because delta depends on whether flipping the bit would add or subtract d.\n\nWait,",
    "### Question:\nYou are given two positive integers n and limit.\nReturn the total number of ways to distribute n candies among 3 children such that no child gets more than limit candies.\n \nExample 1:\n\nInput: n = 5, limit = 2\nOutput: 3\nExplanation: There are 3 ways to distribute 5 candies such that no child gets more than 2 candies: (1, 2, 2), (2, 1, 2) and (2, 2, 1).\n\nExample 2:\n\nInput: n = 3, limit = 3\nOutput: 10\nExplanation: There are 10 ways to distribute 3 candies such that no child gets more than 3 candies: (0, 0, 3), (0, 1, 2), (0, 2, 1), (0, 3, 0), (1, 0, 2), (1, 1, 1), (1, 2, 0), (2, 0, 1), (2, 1, 0) and (3, 0, 0).\n\n \nConstraints:\n\n1 <= n <= 50\n1 <= limit <= 50\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def distributeCandies(self, n: int, limit: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the number of ways to distribute n candies among 3 children such that none of them gets more than the limit. Let's think about how to approach this.\n\nFirst, let me understand the problem. Each child can get between 0 and limit candies, inclusive. The sum of the three children's candies must be exactly n. And I have to count all possible ordered triples (a, b, c) where a + b + c = n, and each of a, b, c is between 0 and limit.\n\nThe examples given: For n=5 and limit=2, the valid triples are permutations of (2,2,1). There are 3 ways. For n=3 and limit=3, there are 10 ways, including all permutations where the sum is 3. \n\nSo the straightforward way is to generate all possible triples and check the conditions, but since n can be up to 50, and 3 children, the total possibilities without constraints would be O(50^3) which is 125,000. Since 50^3 is manageable for a brute force approach. But maybe there's a smarter mathematical way.\n\nBut given that n and limit are up to 50, perhaps a brute force approach is acceptable here. Let me think about the constraints.\n\nBut wait, for n=50, and each child can have up to 50, so for each a from 0 to min(50, 50), then for each b similarly, etc. The triple loops would iterate up to 50 each. The total number of iterations would be 50*50*50 = 125,000. Which is totally acceptable for Python. So perhaps the simplest way is to iterate through all possible a, b, c such that a + b + c = n, and each is within the limit. But wait, how do I generate all a, b, c that sum to n?\n\nAlternatively, perhaps we can compute the total number of solutions (without considering the limit) and then subtract those solutions that violate the limit. That's another approach.\n\nThe standard stars and bars problem: The number of non-negative integer solutions to a + b + c = n is C(n + 2, 2). But this is only when n >=0 and each variable can be any non-negative integer. However, here we have upper bounds on each variable.\n\nSo inclusion-exclusion might be necessary. Let me think.\n\nThe formula for the number of solutions where each variable is at most limit is:\n\nTotal = C(n + 2, 2) - 3*C( (n - (limit + 1)) + 2, 2 ) + 3*C( (n - 2*(limit +1)) + 2, 2 ) - C( (n - 3*(limit+1)) + 2, 2 )\n\nBut only when the terms are valid. Wait, that's inclusion-exclusion. Let me recall the inclusion-exclusion formula for upper bounds.\n\nThe problem is to find the number of non-negative integer solutions to a + b + c = n, where 0 ≤ a, b, c ≤ limit.\n\nThe standard inclusion-exclusion approach would start with the total number of solutions without constraints (which is C(n + 2, 2)), then subtract the cases where at least one variable exceeds the limit, add back those where two variables exceed, etc.\n\nSo, the formula is:\n\nNumber of valid solutions = Total - C(3,1)*A + C(3,2)*B - C(3,3)*C, where:\n\nA is the number of solutions where a > limit (assuming a is the one exceeding), \n\nB where a>limit and b>limit, etc.\n\nC where a, b, c all exceed.\n\nBut for each of these terms, how do we compute them?\n\nFor A: If a exceeds limit, then let a' = a - (limit + 1), which is ≥0. Then the equation becomes a' + (limit +1) + b + c = n → a' + b + c = n - (limit +1). The number of solutions is the number of non-negative solutions to this equation, which is C( (n - (limit +1)) + 2, 2 ), but only if n - (limit +1) ≥ 0. Otherwise, it's zero.\n\nSimilarly, for B: two variables exceed. Let's say a and b. Then a ≥ limit +1, b ≥ limit +1. Let a' = a - (limit+1), b' = b - (limit+1). Then the equation becomes a' + b' + c + (limit+1) + (limit+1) = n → a' + b' + c = n - 2*(limit +1). The number of solutions here is C( (n - 2*(limit +1)) + 2, 2 ), again if the value is ≥0.\n\nSame for three variables: all three exceed. Then the equation becomes a' + b' + c' = n - 3*(limit + 1), and the number of solutions is C(..., 2) if possible.\n\nSo applying inclusion-exclusion:\n\nThe formula is:\n\nnumber of valid solutions = S0 - 3*S1 + 3*S2 - 3*S3,\n\nWait, let me rephrase:\n\nThe inclusion-exclusion formula for three sets is:\n\n|A ∪ B ∪ C| = |A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| | + |A∩B∩C|.\n\nSo the number of solutions where at least one variable exceeds is 3|A| - 3|A∩B| + |A∩B∩C}|. Therefore, the valid solutions are total (without constraints) minus |A ∪ B ∪ C|.\n\nSo the valid solutions = total - |A ∪ B ∪ C| \n\n= total - (3|A| - 3|A∩B| + |A∩B∩C| )) \n\n= total - 3|A| + 3|A∩B| - |A∩B∩C|.\n\nSo the valid solutions is:\n\nC(n + 2, 2) - 3 * C(n - (L+1) + 2, 2) + 3 * C(n - 2*(L+1) + 2, 2) - C(n - 3*(L+1) + 2, 2),\n\nbut only when the arguments for each term are >=0. Otherwise, those terms are zero.\n\nWait, but L here is the limit. Let me denote L as the limit.\n\nBut each term must be considered only when the value inside the combination is >=0. Otherwise, that term is zero.\n\nSo the formula would be:\n\nvalid = stars_and_bars_total(n) \n\n- 3 * stars_and_bars_total(n - (L+1)) \n\n+ 3 * stars_and_bars_total(n - 2*(L+1)) \n\n- stars_and_bars_total(n - 3*(L+1)).\n\nBut stars_and_bars_total(k) is only valid when k >=0. Otherwise, that term is zero.\n\nWait, the stars and bars total here is for the equation a + b + c = k, which requires that k >=0. But if k is negative, then there are zero solutions. So for each term in the inclusion-exclusion formula, we need to calculate the combination if the argument is >=0, else zero.\n\nBut how to compute stars_and_bars_total(k) here? For the original problem, the number of non-negative integer solutions to a + b + c = k is (k + 2) choose 2, but only if k >=0. Otherwise zero.\n\nSo putting it all together, the formula for valid is:\n\nvalid = max(0, comb(n + 2, 2)) \n\n- 3 * max(0, comb(n - (L+1) + 2, 2)) \n\n+ 3 * max(0, comb(n - 2*(L+1) + 2, 2)) \n\n- max(0, comb(n - 3*(L+1) + 2, 2)).\n\nWait, no. Because in inclusion-exclusion, each term is computed only if applicable. For example, the first term in inclusion-exclusion is 3*|A| where |A| is the number of solutions where a exceeds L. Similarly for others.\n\nSo for each term, we compute:\n\nterm1 = 3 * stars_and_bars(m1), where m1 = n - (L+1). But only if m1 >=0.\n\nSimilarly term2 is 3 * stars_and_bars(m2), where m2 = n - 2*(L+1), if m2 >=0, else zero.\n\nterm3 is stars_and_bars(m3), where m3 = n - 3*(L+1), if m3 >=0, else zero.\n\nSo the formula is:\n\nvalid = total_solutions - term1 + term2 - term3 ?\n\nWait, no. Let me recheck:\n\nThe inclusion-exclusion gives:\n\nvalid = total - (3A - 3B + C) ?\n\nNo. Let me re-express:\n\nThe inclusion-exclusion says that the valid solutions are:\n\nTotal - (number of solutions with at least one variable exceeding).\n\nNumber of solutions with at least one variable exceeding is:\n\nSum (each single variable exceeding) - Sum (each pair exceeding) + Sum (all three exceeding).\n\nWhich is:\n\n3*A_single - 3*B_pair + 1*C_triple.\n\nSo the valid solutions is:\n\nTotal - [3*A_single - 3*B_pair + C_triple] \n\n= Total - 3*A_single + 3*B_pair - C_triple.\n\nWhere:\n\nA_single is the number of solutions where a exceeds (which is stars_and_bars for n - (L+1)), same for b and c. Hence multiplied by 3.\n\nB_pair is the number of solutions where a and b exceed, which is stars_and_bars for n - 2*(L+1). Similarly for a and c, and b and c. So multiplied by 3.\n\nC_triple is the number where all three exceed",
    "### Question:\nYou are given a string s and a positive integer k.\nLet vowels and consonants be the number of vowels and consonants in a string.\nA string is beautiful if:\n\nvowels == consonants.\n(vowels * consonants) % k == 0, in other terms the multiplication of vowels and consonants is divisible by k.\n\nReturn the number of non-empty beautiful substrings in the given string s.\nA substring is a contiguous sequence of characters in a string.\nVowel letters in English are 'a', 'e', 'i', 'o', and 'u'.\nConsonant letters in English are every letter except vowels.\n \nExample 1:\n\nInput: s = \"baeyh\", k = 2\nOutput: 2\nExplanation: There are 2 beautiful substrings in the given string.\n- Substring \"baeyh\", vowels = 2 ([\"a\",e\"]), consonants = 2 ([\"y\",\"h\"]).\nYou can see that string \"aeyh\" is beautiful as vowels == consonants and vowels * consonants % k == 0.\n- Substring \"baeyh\", vowels = 2 ([\"a\",e\"]), consonants = 2 ([\"b\",\"y\"]). \nYou can see that string \"baey\" is beautiful as vowels == consonants and vowels * consonants % k == 0.\nIt can be shown that there are only 2 beautiful substrings in the given string.\n\nExample 2:\n\nInput: s = \"abba\", k = 1\nOutput: 3\nExplanation: There are 3 beautiful substrings in the given string.\n- Substring \"abba\", vowels = 1 ([\"a\"]), consonants = 1 ([\"b\"]). \n- Substring \"abba\", vowels = 1 ([\"a\"]), consonants = 1 ([\"b\"]).\n- Substring \"abba\", vowels = 2 ([\"a\",\"a\"]), consonants = 2 ([\"b\",\"b\"]).\nIt can be shown that there are only 3 beautiful substrings in the given string.\n\nExample 3:\n\nInput: s = \"bcdf\", k = 1\nOutput: 0\nExplanation: There are no beautiful substrings in the given string.\n\n \nConstraints:\n\n1 <= s.length <= 1000\n1 <= k <= 1000\ns consists of only English lowercase letters.\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def beautifulSubstrings(self, s: str, k: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to find the number of beautiful substrings in a given string s with a certain k. The conditions for a substring being beautiful are two: vowels equals consonants, and the product of vowels and consonants is divisible by k. \n\nFirst, let me think about what these conditions imply. For vowels to equal consonants, the total length of the substring must be even. Because vowels + consonants = total, and if vowels equals consonants, then total is 2*vowels, so even. So any valid substring must have even length. That's an important point. So maybe I can limit my search to substrings with even lengths.\n\nSo the first step is to check all possible substrings, but only those with even lengths. For each such substring, count vowels and consonants, check if they are equal in count, and then check if their product is divisible by k. But checking all substrings directly might be computationally heavy, but since the constraints are up to 1000 characters, maybe a brute-force approach with optimizations is feasible.\n\nLet me think about the brute-force approach. For each possible substring (start and end indices), check if it meets the criteria. But how to compute vowels and consonants efficiently?\n\nLet me think about precomputing some prefix sums. For example, for each index i, compute the cumulative number of vowels up to i. Then, for any substring from index i to j, the number of vowels is prefix[j] - prefix[i], and consonants is (length of substring) - vowels. But since vowels and consonants must be equal, that implies that vowels = consonants = (j-i+1)/2. And vowels must equal consonants, so the total vowels in the substring must be exactly half the length of the substring.\n\nSo for a substring from i (inclusive) to j (inclusive), which has length L = j - i + 1. Since L must be even, we can iterate over all possible even lengths. Then for each possible even length, check all starting positions, compute vowels and consonants.\n\nSo let's think about steps:\n\n1. Precompute a prefix sum array for vowels. Let's say prefix_vowel[i] is the number of vowels in s[0..i-1], so that prefix_vowel[0] = 0, prefix_vowel[1] is 1 if s[0] is vowel else 0, etc. Then for substring s[a..b] (0-based, inclusive), the number of vowels is prefix_vowel[b+1] - prefix_vowel[a]. The length of this substring is (b - a + 1). \n\nSo for the substring to be beautiful, two conditions:\n\na) vowels == consonants => vowels = (length)/2. Since length is even, let's say length is 2m. So vowels must be exactly m.\n\nb) vowels * consonants = m * m. This product must be divisible by k. So m² mod k must be 0. Therefore, k must divide m².\n\nSo for each substring of even length, check if vowels count is exactly half the length (i.e., m), and m² is divisible by k.\n\nSo how to approach this?\n\nFirst, precompute the prefix vowels array.\n\nThen, for all possible substrings (i, j), check if length is even. If yes, compute m = length / 2. Compute vowels in substring. If vowels is m, then check if m*m is divisible by k. If yes, then count this substring.\n\nBut checking all substrings would be O(n²) time. For n=1000, that's 1e6 operations. For each such substring, checking m*m %k is O(1). But for n=1e3, 1e6 is acceptable. Let's see:\n\nn=1000, the number of possible substrings is n*(n+1)/2 ≈ 5e5. But considering only even-length substrings, it's roughly half that, about 250,000. But even if it's 5e5, for each substring, the checks are O(1), so total time would be acceptable.\n\nSo the plan is:\n\n- Precompute prefix_vowel array.\n\n- Iterate over all possible substrings. For each:\n\n   - Check if length is even.\n\n   - Compute vowels count.\n\n   - Check if vowels count is m = length/2.\n\n   - If yes, check if m² mod k == 0. If yes, increment count.\n\nSo, let's think about implementing this.\n\nFirst, implement the prefix sum array. Let's say s is given as input. For example, for s = \"baeyh\", the prefix_vowel array would be:\n\nprefix_vowel[0] = 0\n\nprefix_vowel[1] = 1 (since first character is 'b' no, wait. Wait, s[0] is 'b'? Let me think. Oh, the code needs to process the string properly. For example, for each character in s, check if it is a vowel.\n\nLet me code the helper function to check if a character is a vowel. Then build prefix_vowel array.\n\nOnce that is done, for each possible substring, compute the vowels count. For example, for substring starting at index a and ending at index b (0-based), where a <= b. The length is (b - a + 1). So for the length to be even, (b - a + 1) must be even. Let me think for a moment.\n\nNow, how to iterate all possible substrings? For each possible start and end. Let's think in code:\n\nfor i in range(len(s)):\n\n    for j in range(i+1, len(s)):\n\n        length = j - i + 1 ?\n\nWait, for indexes i and j, where i <= j. The substring s[i...j] (inclusive) has length j - i + 1. So for example, i=0, j=0: length 1.\n\nSo for even lengths, j - i + 1 must be even. So j - i + 1 = even => j -i is odd. So i and j have opposite parity? Not sure, but perhaps easier to check for the length.\n\nSo in code:\n\nfor start in 0 to n-1:\n\n    for end in start to n-1:\n\n        length = end - start + 1\n\n        if length is even:\n\n            vowels_count = prefix_vowel[end+1] - prefix_vowel[start]\n\n            m = length / 2\n\n            if vowels_count == m:\n\n                if (m * m) % k == 0:\n\n                    increment count\n\nSo that's the plan. Now, what about the case when m is zero? Well, since length is 0? No, since start <= end, length is at least 1, and when length is even, the minimum is 2. So m is at least 1.\n\nWait, for length 2, m is 1. So vowels_count must be 1, consonants is 1. So product is 1. Then check if 1 mod k is zero. For example, if k is 2, then 1 mod 2 is 1, no. So not counted.\n\nBut in example 1, let's see. For the substring \"aeyh\" which is from index 1 to 4 (assuming 0-based). The original string is \"baeyh\". So indexes 0: 'b', 1:'a', 2:'e', 3:'y', 4:'h'. So start is 1, end is 4: length 4. m is 2. vowels_count is prefix[5] (since end+1 is 4+1=5) minus prefix[1]. Let's compute prefix_vowel for example.\n\nLet me take example 1: s = \"baeyh\".\n\nEach character:\n\nb: consonant (0)\n\na: vowel (1)\n\ne: vowel (2)\n\ny: consonant (2)\n\n",
    "### Question:\nYou are given a 0-indexed array nums consisting of positive integers.\nA partition of an array into one or more contiguous subarrays is called good if no two subarrays contain the same number.\nReturn the total number of good partitions of nums.\nSince the answer may be large, return it modulo 10^9 + 7.\n \nExample 1:\n\nInput: nums = [1,2,3,4]\nOutput: 8\nExplanation: The 8 possible good partitions are: ([1], [2], [3], [4]), ([1], [2], [3,4]), ([1], [2,3], [4]), ([1], [2,3,4]), ([1,2], [3], [4]), ([1,2], [3,4]), ([1,2,3], [4]), and ([1,2,3,4]).\n\nExample 2:\n\nInput: nums = [1,1,1,1]\nOutput: 1\nExplanation: The only possible good partition is: ([1,1,1,1]).\n\nExample 3:\n\nInput: nums = [1,2,1,3]\nOutput: 2\nExplanation: The 2 possible good partitions are: ([1,2,1], [3]) and ([1,2,1,3]).\n\n \nConstraints:\n\n1 <= nums.length <= 10^5\n1 <= nums[i] <= 10^9\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def numberOfGoodPartitions(self, nums: List[int]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to find the number of good partitions of the array nums. A good partition is when each subarray in the partition doesn't share any numbers with another. And since the answer can be big, I have to return it modulo 1e9+7.\n\nHmm. Let me think about what defines a good partition. Each subarray in the partition must have unique elements compared to the others. Wait, no, wait. Wait the problem says no two subarrays can contain the same number. So for example, if a number appears in more than one subarray of the partition, that's invalid. So each number in the original array can only appear in one subarray of the partition. Because if a number appears multiple times in the original array, you can't have it in two different subarrays of the partition.\n\nSo the key here is to figure out how to arrange the cuts in the array such that for each number, all its occurrences are contained within a single subarray of the partition. Because otherwise, if there's a number that appears in two different subarrays, that's not allowed.\n\nSo the problem reduces to determining the points where such cuts can be made. For example, if some numbers appear multiple times, their first and last occurrence must be in the same subarray. Therefore, perhaps we need to find intervals where each number's first and last occurrence are considered, and then find overlapping intervals and determine where to split.\n\nLet me think of an approach. For each number, we can note its first and last occurrence in the array. For example, if a number x appears at positions i and j (i < j), then in any valid partition, all the occurrences of x must be in a single subarray. Therefore, any cut must happen before the first occurrence of x or after the last occurrence of x. So, for example, if x's first occurrence is at index i and last at j, then between i and j, we can't have a cut that splits this region. So the idea is that overlapping intervals of first and last occurrences may impose constraints on where we can split the array.\n\nThis reminds me of the problem where you have to find the number of ways to split the array into intervals where certain conditions are met. For example, the problem where you have to make sure that each element is in exactly one interval, and you have to count the possible splits. In such cases, the approach is to merge overlapping intervals and then calculate based on the gaps between intervals.\n\nLet me think. Let's take example 1: [1,2,3,4]. Each element is unique. So their first and last occurrence is the same. For each element, their interval is [i,i]. So the constraints here are that for each element, any split must not split between their occurrence. But since they are single elements, perhaps the splits can be made between any positions. How many splits? Well, for n elements, there are n-1 possible positions between elements. For each such position, you can choose to split or not. However, the total number of partitions would depend on how many splits you make. For example, if you make all possible splits, you get n partitions. But the problem counts all possible valid partitions. However, in this case, since each element's interval is [i,i], any splits are allowed. So the total number of possible good partitions would be 2^(k), where k is the number of gaps between elements. Because for each gap, you can choose to split or not. Wait, but no. Because if you make splits in some gaps but not others, the number of partitions depends on which splits you make. The total number of possible partitions is 2^{n-1} }? No, that's not possible. For example, for n=4, example 1 says 8. Which is 2^3. So yes, 2^{n-1} possible ways. Because for each of the n-1 gaps between elements, you can choose to split or not. But the total number of possible partitions is indeed 2^{n-1} because each choice of splits defines a unique partition. For example, each time you make splits in certain positions. However, in our problem, not all splits are allowed. Because when there are overlapping intervals.\n\nSo the main idea is that, considering all the intervals (for each element, the interval between first and last occurrence), the splits can only be made at the end of such intervals. Because once you have a interval that must be contained in a single subarray, the splits can only be made after certain points.\n\nLet me think again. Suppose that after merging overlapping intervals, we have some non-overlapping intervals. For example, if the intervals for different elements don't overlap. Then, between these merged intervals, you can choose to split or not. But how?\n\nWait, let's think of merging all the intervals. For example, if some elements' intervals overlap, their merged interval is the minimal interval that contains all their individual intervals. For example, consider [1,2,1,3]. The first occurrence of 1 is 0, last is 2. For 2, it's [1,1], for 3 it's [3,3]. So the merged intervals would be [0,2], [1,1], [3,3]. Wait, but how do we compute them? Because the merged intervals must be such that any overlapping or contained intervals are merged. For example, since the interval [1,1] is inside [0,2], perhaps that doesn't add any new constraints. So the key is that the merged intervals are non-overlapping and cover the entire array. Once we have these merged intervals, the splits can only be made between them. Because each merged interval must be in a single subarray. Therefore, between two merged intervals, we can choose to split or not. But how many splits are allowed?\n\nFor example, if there are m merged intervals, then between them there are m-1 possible positions to split or not. But the number of possible splits would be 2^{m-1} possible ways. Because between each pair of merged intervals, we can decide to split or not. Because once the merged intervals are determined, each must be part of exactly one subarray. So if the merged intervals are non-overlapping and consecutive, then between any two consecutive merged intervals, we can make a split. Each such split can be chosen independently. Therefore, the answer is 2^(number of merged gaps), where the number is m-1, with m being the number of merged intervals.\n\nLet me test this idea with the examples.\n\nExample 1: [1,2,3,4]. Each element's interval is [i,i]. Merged intervals would be [0,0], [1,1], [2,2], [3,3]. Since they are all consecutive, the merged intervals are four. So the number of possible splits is 2^(4-1) = 8. Which matches the example.\n\nExample 2: [1,1,1,1]. The interval for 1 is [0,3]. So merged interval is just one interval. So number of splits is 2^0 = 1. Which matches the example.\n\nExample 3: [1,2,1,3]. The merged intervals are [0,2], [3,3]. So two intervals. So 2^(2-1) = 2. Which matches the example. Yes! So this seems to work.\n\nSo the problem reduces to:\n\n1. For each element in nums, find its first and last occurrence.\n\n2. Merge all overlapping intervals. The order of merging isn't important if we merge all overlapping ones.\n\n3. The number of merged intervals is m. The answer is",
    "### Question:\nYou are given a 0-indexed integer array batteryPercentages having length n, denoting the battery percentages of n 0-indexed devices.\nYour task is to test each device i in order from 0 to n - 1, by performing the following test operations:\n\nIf batteryPercentages[i] is greater than 0:\n\n\t\nIncrement the count of tested devices.\nDecrease the battery percentage of all devices with indices j in the range [i + 1, n - 1] by 1, ensuring their battery percentage never goes below 0, i.e, batteryPercentages[j] = max(0, batteryPercentages[j] - 1).\nMove to the next device.\n\n\nOtherwise, move to the next device without performing any test.\n\nReturn an integer denoting the number of devices that will be tested after performing the test operations in order.\n \nExample 1:\n\nInput: batteryPercentages = [1,1,2,1,3]\nOutput: 3\nExplanation: Performing the test operations in order starting from device 0:\nAt device 0, batteryPercentages[0] > 0, so there is now 1 tested device, and batteryPercentages becomes [1,0,1,0,2].\nAt device 1, batteryPercentages[1] == 0, so we move to the next device without testing.\nAt device 2, batteryPercentages[2] > 0, so there are now 2 tested devices, and batteryPercentages becomes [1,0,1,0,1].\nAt device 3, batteryPercentages[3] == 0, so we move to the next device without testing.\nAt device 4, batteryPercentages[4] > 0, so there are now 3 tested devices, and batteryPercentages stays the same.\nSo, the answer is 3.\n\nExample 2:\n\nInput: batteryPercentages = [0,1,2]\nOutput: 2\nExplanation: Performing the test operations in order starting from device 0:\nAt device 0, batteryPercentages[0] == 0, so we move to the next device without testing.\nAt device 1, batteryPercentages[1] > 0, so there is now 1 tested device, and batteryPercentages becomes [0,1,1].\nAt device 2, batteryPercentages[2] > 0, so there are now 2 tested devices, and batteryPercentages stays the same.\nSo, the answer is 2.\n\n \nConstraints:\n\n1 <= n == batteryPercentages.length <= 100 \n0 <= batteryPercentages[i] <= 100\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def countTestedDevices(self, batteryPercentages: List[int]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. The problem is about counting how many devices are tested based on certain conditions. The battery percentages of devices are given, and for each device in order, if its battery is more than 0, we test it, increment the count, and then decrease all devices after it by 1 (but not below zero). But doing this naively for each step might be inefficient, especially because for each tested device, we have to update all subsequent devices. But since the constraints are small (n up to 100), maybe a straightforward approach is acceptable.\n\nWait, but let's think about how to model the battery percentages. Let's take example 1: [1,1,2,1,3]. When we test device 0, all devices after it (indices 1-4) get decreased by 1. So the array becomes [1,0,1,0,2]. Then when processing device 1, it's 0, so skip. Device 2 is 1, which is >0. After testing, devices after 2 (indices 3 and 4) get decreased by 1. So index 3 becomes 0-1=0, and index 4 is 2-1=1. Then device 3 is 0, skip. Device 4 is 1, test. So total 3.\n\nThe straightforward way would be to simulate each step. For each device i from 0 to n-1:\n\n- if battery[i] > 0:\n\n   - increment count\n\n   - for j from i+1 to n-1:\n\n      - battery[j] = max(0, battery[j] - 1)\n\nBut since for each tested device, we have to iterate through the rest of the array, the time complexity would be O(n^2), which for n=100 is acceptable. So maybe this approach is feasible.\n\nSo implementing this simulation would be straightforward. Let's think about how to code this.\n\nThe function is given as a method in a class. The input is batteryPercentages as a list of integers. We need to return the count.\n\nSo, in Python, we can loop through each index i from 0 to len(batteryPercentages)-1. For each i, check if batteryPercentages[i] is > 0. If yes, increment the count. Then for all j > i, decrease by 1, but not below 0.\n\nLet me try to code this.\n\nInitialize count to 0. Then for i in range(len(battery)):\n\nif battery[i] > 0:\n\n   count +=1\n\n   for j in range(i+1, len(battery)):\n\n       battery[j] = max(0, battery[j] -1)\n\nelse:\n\n   continue\n\nReturn count.\n\nBut wait, in Python, lists are mutable, so modifying the input list is okay here. Because for each step, the changes to the list affect the subsequent steps. So this approach would work.\n\nTesting example 1:\n\nOriginal list is [1,1,2,1,3]. Let's walk through each step.\n\ni=0: battery[0] is 1>0. count becomes 1. Then for j=1 to 4:\n\nj=1: 1-1=0. So [1,0, ...]\n\nj=2: 2-1=1. So battery[2] becomes 1.\n\nj=3: 1-1=0.\n\nj=4: 3-1=2. So battery becomes [1,0,1,0,2]. Then next i=1. battery[1] is 0, skip. i=2: battery[2] is 1>0, count becomes 2. Then for j=3 and 4:\n\nj=3: 0-1 (but set to 0). j=4: 2-1=1. So battery becomes [1,0,1,0,1]. Then i=3, battery is 0, skip. i=4, battery is 1>0, count becomes 3. Then no j's after. So returns 3. Correct.\n\nAnother example, input [0,1,2]. Let's see:\n\ni=0: battery[0] is 0, skip. i=1: battery is 1>0. count becomes 1. Then j=2: 2-1=1. So battery becomes [0,1,1]. i=2: battery is 1>0, count becomes 2. Then no j's after. So return 2. Correct.\n\nAnother test case. Let's say input [2, 2, 2]. Let's see:\n\ni=0: battery is 2>0. count=1. Then subtract 1 from j=1 and 2. They become 1 and 1.\n\ni=1: battery is 1>0. count=2. Subtract 1 from j=2: 1-1=0. So battery[j] is 0.\n\ni=2: battery is 0. so skip. total count is 2? Or wait:\n\nWait after i=1, the battery at j=2 is 1 (original was 1, subtract 1 becomes 0). So at i=2, battery is 0. So total count is 2. But wait, original input [2,2,2], after first step (i=0):\n\nbattery becomes [2, 1, 1]. Then i=1, which is 1>0, count becomes 2. Subtract 1 from j=2: 1-1=0. So battery[j] is 0. Then i=2: battery is 0. So total count is 2. But wait, the initial battery for i=0 was 2. When we test it, we do the subtraction. Then for i=1, after subtraction, but it's still positive. So yes, count is 2. But wait, what about the first device's own battery? Does it get modified? No. Because when you test device i, you decrease the battery of devices after i, not including i itself. So in this example, the first device's battery remains 2. Then when i=1, battery is 2-1=1 (from the initial 2, minus 1 because of the first step). Then tested, count increases, and then subtract 1 from j=2. So yes, the count is 2.\n\nAnother test case: [3]. So n=1. i=0: battery is 3>0, count becomes 1. Return 1. Correct.\n\nAnother case: [0]. Return 0.\n\nAnother case: [5,0,0]. Let's see. i=0, battery is 5>0. count becomes 1. Subtract 1 from j=1 and 2. They become 0-1 (but 0) and 0-1 (but 0). So battery is [5,0,0]. Then i=1: battery is 0. skip. i=2: battery is 0. So total count 1.\n\nYes. So the simulation seems to work.\n\nNow, considering the constraints. For n=100, the worst case is O(n^2) = 10^4 operations, which is acceptable.\n\nTherefore, the approach is to simulate the process as described.\n\nSo coding this:\n\nIn the function, create a variable count = 0.\n\nLoop over each i from 0 to len(batteryPercentages) - 1:\n\n   if batteryPercentages[i] > 0:\n\n      count += 1\n\n      for j in range(i+1, len(batteryPercentages)):\n\n          batteryPercentages[j] = max(0, batteryPercentages[j] - 1)\n\nReturn count.\n\nSo this should work.\n\nLet me check for possible errors.\n\nWait, what about when batteryPercentages[i] is 0 but after some previous subtractions, it's still 0. Like in example 2, the initial battery is [0,1,2]. For device 0, it's 0. Then when device 1 is processed, the battery of device 2 is 2-1=1. Then device 2 is processed, which is 1>0, so count increases by 1. So code seems to handle that.\n\nAnother case where after subtraction, some devices in the future have their battery reduced. But since we process them in order, and for each tested device i, we reduce all j > i, then the next devices are processed with the updated values.\n\nYes, the code should handle that.\n\nTherefore, the code should be correct. Let me test the examples.\n\n",
    "### Question:\nYou are given a string word and an integer k.\nA substring s of word is complete if:\n\nEach character in s occurs exactly k times.\nThe difference between two adjacent characters is at most 2. That is, for any two adjacent characters c1 and c2 in s, the absolute difference in their positions in the alphabet is at most 2.\n\nReturn the number of complete substrings of word.\nA substring is a non-empty contiguous sequence of characters in a string.\n \nExample 1:\n\nInput: word = \"igigee\", k = 2\nOutput: 3\nExplanation: The complete substrings where each character appears exactly twice and the difference between adjacent characters is at most 2 are: igigee, igigee, igigee.\n\nExample 2:\n\nInput: word = \"aaabbbccc\", k = 3\nOutput: 6\nExplanation: The complete substrings where each character appears exactly three times and the difference between adjacent characters is at most 2 are: aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc.\n\n \nConstraints:\n\n1 <= word.length <= 10^5\nword consists only of lowercase English letters.\n1 <= k <= word.length\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def countCompleteSubstrings(self, word: str, k: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the number of complete substrings in a given word based on certain conditions. Let me think through how to approach this.\n\nFirst, the problem defines a complete substring as one where each character occurs exactly k times, and the adjacent characters differ by at most 2 in their alphabet positions. So for any substring I check, both conditions must be satisfied.\n\nLet me break down the two conditions:\n\n1. Each character in the substring appears exactly k times. So the length of the substring must be a multiple of k. Let's say if there are m distinct characters in the substring, then the total length is m * k.\n\n2. The adjacent characters differ by at most 2. For example, 'a' next to 'c' is okay (difference 2), but 'a' next to 'd' would be difference 3 which is not allowed.\n\nSo how can I efficiently check all possible substrings of the word for these conditions?\n\nThe brute-force approach would be to check every possible substring, but given the constraints (word length up to 1e5), that's going to be too slow. So I need a smarter approach.\n\nLet me think about possible optimizations.\n\nObservation 1: The second condition (adjacent characters differ by at most 2) implies that the substring is such that the characters are in a sequence where each consecutive character is very close. For example, if the substring contains 'a', then the next characters can only be 'a', 'b', or 'c'. So perhaps the valid substrings are those where the characters form a \"smooth\" sequence.\n\nThis suggests that the entire substring's characters must be part of a consecutive group in the alphabet. For example, maybe a substring can have 'a' and 'b' but not 'c', or 'b' and 'c' but not 'a', etc. However, this isn't straightforward because the order in the substring isn't necessarily sorted. For example, \"acb\" would have adjacent differences of 2 between a and c (if adjacent), but then between c and b also 1. Wait, but in the substring, the order of characters matters. So for example, the substring could be \"acb\", which has adjacent differences of 2 and 1. So the characters can be in any order as long as adjacent ones are close.\n\nBut how can I check this efficiently for all possible substrings?\n\nAnother thought: since the adjacent characters must differ by at most 2, perhaps the entire substring's characters can only span a small range. For example, if the substring contains characters from 'a' to 'a + m', where m is small (like 2 or 3). For example, a substring can't have both 'a' and 'd' unless there's a 'b' or 'c' in between, but even then, the arrangement of the letters in the substring must ensure that adjacent letters are within 2.\n\nThis seems complex. Maybe there's a way to split the original word into segments where each segment can have valid substrings. For example, whenever there's a point where consecutive characters differ by more than 2, we can split the word here. Because any valid substring can't cross such a split.\n\nFor example, consider the word \"abcxdef\". The 'c' and 'x' differ by 23, which is way more than 2. So any valid substring can't include both 'c' and 'x'. Therefore, we can split the original word into parts like \"abc\", \"x\", \"def\", and process each part separately. Then, for each of these segments, we look for valid substrings.\n\nThis seems promising. Because splitting the word into these segments would reduce the problem into smaller parts, and for each part, when checking substrings, we don't have to worry about the second condition (adjacent differences) except for the order within the substring. Wait, no. Because even within a segment like \"abc\", a substring like \"acb\" is allowed as long as adjacent characters differ by at most 2. For example, in \"acb\", the first two characters (a and c) differ by 2, and the next two (c and b) differ by 1. So that's valid. But how do I ensure that when considering substrings in the split segments?\n\nThe idea is that by splitting the original word into segments where consecutive characters in the original word differ by more than 2, then any valid substring must lie entirely within one of these segments. Because if a substring crosses a split point, then the original word has two adjacent characters (at the split) that differ by more than 2. But since the substring is contiguous, it would include those two characters as adjacent or nearby. However, in the substring, they may not be adjacent. For example, if the original word is \"ab...x...yz\", and the split is between ... and x, then a substring could start before the split and end after. However, in that substring, there might be two characters that were split, but their positions in the substring may not be adjacent. But the problem requires that all adjacent characters in the substring have difference <= 2. So even if the original word's characters at the split are far apart, as long as the substring's adjacent characters are okay, it's allowed.\n\nWait, but if the original word has two consecutive characters (say, at positions i and i+1) that differ by more than 2, then any substring that includes both these characters must have them as adjacent in the substring (since the substring is contiguous). Therefore, such a substring would automatically be invalid. Therefore, any valid substring cannot contain two consecutive characters in the original word that differ by more than 2. Therefore, the entire valid substring must lie within a segment where all consecutive characters in the original word differ by at most 2. Therefore, splitting the original word into such segments would allow us to process each segment independently.\n\nFor example, in the original word, we can split it into groups where for each group, every consecutive pair of characters in the original word has a difference of at most 2. Then, any valid substring must be entirely contained within one of these groups.\n\nThis is a crucial insight. So first, I need to split the original word into these groups. For example, if the word is \"igigee\", let's check consecutive characters:\n\nLet me take example 1: word is \"igigee\". Let's check consecutive characters:\n\ni (105), g (103): difference is 2 → okay.\ng and i: difference 2 → okay.\ni and g: same as before.\ng and e: difference is 2 (g is 103, e is 101 → 2).\ne and e: 0 → okay.\n\nSo the entire word is one group. So no splits needed. So all valid substrings are within this group.\n\nAnother example: word is \"aaabbbccc\" from example 2. The consecutive characters are a and a (diff 0), a and a (0), a and b (1), then b and b (0), etc. All consecutive differences are <= 2. So again, the entire word is one group. Hence, process as a single group.\n\nBut if there's a word like \"abcf\", then consecutive differences are a-b (1), b-c (1), c-f (3). So the split would occur between c and f. So ",
    "### Question:\nYou are given a 0-indexed string s and an integer k.\nYou are to perform the following partitioning operations until s is empty:\n\nChoose the longest prefix of s containing at most k distinct characters.\nDelete the prefix from s and increase the number of partitions by one. The remaining characters (if any) in s maintain their initial order.\n\nBefore the operations, you are allowed to change at most one index in s to another lowercase English letter.\nReturn an integer denoting the maximum number of resulting partitions after the operations by optimally choosing at most one index to change.\n \nExample 1:\n\nInput: s = \"accca\", k = 2\nOutput: 3\nExplanation: In this example, to maximize the number of resulting partitions, s[2] can be changed to 'b'.\ns becomes \"acbca\".\nThe operations can now be performed as follows until s becomes empty:\n- Choose the longest prefix containing at most 2 distinct characters, \"acbca\".\n- Delete the prefix, and s becomes \"bca\". The number of partitions is now 1.\n- Choose the longest prefix containing at most 2 distinct characters, \"bca\".\n- Delete the prefix, and s becomes \"a\". The number of partitions is now 2.\n- Choose the longest prefix containing at most 2 distinct characters, \"a\".\n- Delete the prefix, and s becomes empty. The number of partitions is now 3.\nHence, the answer is 3.\nIt can be shown that it is not possible to obtain more than 3 partitions.\nExample 2:\n\nInput: s = \"aabaab\", k = 3\nOutput: 1\nExplanation: In this example, to maximize the number of resulting partitions we can leave s as it is.\nThe operations can now be performed as follows until s becomes empty: \n- Choose the longest prefix containing at most 3 distinct characters, \"aabaab\".\n- Delete the prefix, and s becomes empty. The number of partitions becomes 1. \nHence, the answer is 1. \nIt can be shown that it is not possible to obtain more than 1 partition.\n\nExample 3:\n\nInput: s = \"xxyz\", k = 1\nOutput: 4\nExplanation: In this example, to maximize the number of resulting partitions, s[1] can be changed to 'a'.\ns becomes \"xayz\".\nThe operations can now be performed as follows until s becomes empty:\n- Choose the longest prefix containing at most 1 distinct character, \"xayz\".\n- Delete the prefix, and s becomes \"ayz\". The number of partitions is now 1.\n- Choose the longest prefix containing at most 1 distinct character, \"ayz\".\n- Delete the prefix, and s becomes \"yz\". The number of partitions is now 2.\n- Choose the longest prefix containing at most 1 distinct character, \"yz\".\n- Delete the prefix, and s becomes \"z\". The number of partitions is now 3.\n- Choose the longest prefix containing at most 1 distinct character, \"z\".\n- Delete the prefix, and s becomes empty. The number of partitions is now 4.\nHence, the answer is 4.\nIt can be shown that it is not possible to obtain more than 4 partitions.\n\n \nConstraints:\n\n1 <= s.length <= 10^4\ns consists only of lowercase English letters.\n1 <= k <= 26\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def maxPartitionsAfterOperations(self, s: str, k: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the maximum number of partitions possible by changing at most one character in the string. Let's think about how to approach this.\n\nThe problem says that in each step, we take the longest possible prefix with at most k distinct characters. Then delete it and count a partition. We can change one character in the original string before starting. The goal is to maximize the number of partitions.\n\nFirst, I need to understand how the partitioning works without any changes. Let's think about the standard way to compute the number of partitions. For example, for a given string, we process it greedily by taking the longest prefix with up to k distinct characters each time.\n\nSo, for example, the standard approach for the original string would be something like:\n\nInitialize start of partition. For each character, track the distinct count. Once it exceeds k, split there. Then reset and continue with the remaining string.\n\nBut how to model this? Maybe we can precompute the partitions for the original string, and then see how changing a single character can affect the splits.\n\nBut since changing a character can affect multiple splits, we need to find the optimal place to change a character such that the total number of partitions increases the most.\n\nLet me think about the example given. In the first example, changing s[2] allows splitting into three parts. The original string was \"accca\" which would split as follows: the first split would be 'a' (only 1 distinct), then 'c'... Wait no. Let me recheck. Original s is \"accca\", k=2. Without changing, the first split would be the entire string? Because let's see:\n\nOriginal partitions without change:\n\nStart with 'a', then add 'c' (distinct count is 2). Then add another 'c' (still 2), another 'c' (same), then 'a' (still 2). So the entire string is one partition. But after changing one character, they split into three.\n\nSo the idea is that by changing a character, we can create more splits. The challenge is to find where to change to maximize this.\n\nNow, how to model this?\n\nFirst, I need to find for each position in the string, what would be the effect of changing that character to another character. But trying all possibilities for each position would be computationally expensive, especially for a string of length up to 1e4.\n\nBut since we can only change one character, perhaps there's a way to analyze each possible modification and compute how it affects the number of partitions.\n\nBut how to compute the number of partitions efficiently for each possible modification?\n\nAn alternative approach is to first compute the original number of partitions. Then, find if changing any single character can increase this number. The maximum between the original and these possibilities is the answer.\n\nSo the plan is:\n\n1. Compute the original number of partitions (without any changes).\n2. For each position i in the string, consider changing s[i] to every possible character (a-z except original s[i]), and compute how that affects the partitions. Take the maximum over all these possibilities.\n3. Return the maximum between the original count and the best possible modified count.\n\nBut step 2 is O(n * 26 * ...) which might be too slow for n=1e4. So we need an efficient way to evaluate each possible modification.\n\nWait, but for each position, changing it to a different character could affect the splits in some way. How can we model this efficiently?\n\nAlternatively, perhaps for each position, there's a way to determine the optimal character to change it to, which would allow us to calculate the maximum possible increase in partitions for that position.\n\nBut how to compute the partitions when a single character is modified?\n\nLet me think about how the partitioning works. The partitioning is done greedily. Each time, take the longest possible prefix with up to k distinct characters. So for each split, we need to process the string step by step, splitting when adding another character would exceed k.\n\nSo, modifying a character in the string can influence the splits in two ways:\n\n- It could allow a split to occur earlier, thereby creating more splits.\n- Or it could allow splits that were previously blocked by having too many distinct characters.\n\nBut how to model this for each possible modification?\n\nAn alternative idea is to simulate for each possible modification the new partition count. However, for n=1e4 and 26 letters, this would be O(n*26) simulations, each of which takes O(n) time. That would be O(n²*26) which is 1e4 * 1e4 * 26 = 2.6e9 operations. That's way too slow.\n\nSo this brute-force approach is not feasible. We need a smarter way.\n\nSo perhaps we need to find for each position, the optimal way to change it to maximize the partitions, using some pre-processing.\n\nLet me think about how the original partitioning works. Let's first compute the original partitions. For example, for the original string, we ca",
    "### Question:\nYou are given a 0-indexed string s, a string a, a string b, and an integer k.\nAn index i is beautiful if:\n\n0 <= i <= s.length - a.length\ns[i..(i + a.length - 1)] == a\nThere exists an index j such that:\n\t\n0 <= j <= s.length - b.length\ns[j..(j + b.length - 1)] == b\n|j - i| <= k\n\n\n\nReturn the array that contains beautiful indices in sorted order from smallest to largest.\n \nExample 1:\n\nInput: s = \"isawsquirrelnearmysquirrelhouseohmy\", a = \"my\", b = \"squirrel\", k = 15\nOutput: [16,33]\nExplanation: There are 2 beautiful indices: [16,33].\n- The index 16 is beautiful as s[16..17] == \"my\" and there exists an index 4 with s[4..11] == \"squirrel\" and |16 - 4| <= 15.\n- The index 33 is beautiful as s[33..34] == \"my\" and there exists an index 18 with s[18..25] == \"squirrel\" and |33 - 18| <= 15.\nThus we return [16,33] as the result.\n\nExample 2:\n\nInput: s = \"abcd\", a = \"a\", b = \"a\", k = 4\nOutput: [0]\nExplanation: There is 1 beautiful index: [0].\n- The index 0 is beautiful as s[0..0] == \"a\" and there exists an index 0 with s[0..0] == \"a\" and |0 - 0| <= 4.\nThus we return [0] as the result.\n\n \nConstraints:\n\n1 <= k <= s.length <= 10^5\n1 <= a.length, b.length <= 10\ns, a, and b contain only lowercase English letters.\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def beautifulIndices(self, s: str, a: str, b: str, k: int) -> List[int]:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to find all the beautiful indices in the string s. The definition of a beautiful index i is that the substring starting at i matching a, and there exists some j where the substring starting at j matches b, and the absolute difference between j and i is at most k. And then return those i's sorted.\n\nFirst, I need to find all occurrences of a in s, and all occurrences of b in s. Then, for each occurrence of a (let's call it i), check if there's at least one occurrence of b (j) such that |j - i| <=k. And collect all such i's.\n\nBut how to do this efficiently?\n\nLet me think. The straightforward way would be:\n\n1. Find all indices for a (list_a)\n2. Find all indices for b (list_b)\n3. For each i in list_a, check if there exists a j in list_b such that |i - j| <=k.\n\nBut if list_b is big, checking for each i and every j in list_b could be O(N*M), which might be too slow if both lists are large. Since s can be up to 1e5 in length, but a and b can be as short as 1. For example, if a and b are both single characters, then list_a and list_b could be O(1e5) each. Then checking each pair would be O(1e10), which is way too slow.\n\nSo we need a more efficient way to check for each i in list_a whether there's a j in list_b within [i -k, i +k].\n\nHow can we do that?\n\nSince list_b is a list of indices where the substring b occurs, and assuming they are sorted (since we find them in order), for each i in list_a, we can perform a binary search on list_b to see if there's any j in the range [i -k, i +k]. \n\nYes! Because list_b is sorted (since we find occurrences in order from left to right), for each i, we can check if there's any element in list_b that is between (i -k) and (i +k). How to do that with binary search?\n\nFor example, for a given i, the earliest possible j is i -k, and the latest possible j is i +k. So in the sorted list_b, if there's any element >= (i -k) and <= (i +k), then i is beautiful.\n\nSo for each i in list_a, we can find the first element in list_b that is >= (i -k). Then check if that element is <= (i +k). If yes, then there exists such a j.\n\nWait, but maybe there's a j before that element? No, because if the first element >= i -k is <= i +k, then that's sufficient. Because once you have even one element in that interval, the answer is yes. So we can stop once we find the first occurrence in that range. Alternatively, perhaps using bisect_left to find the insertion point, then check if the element at that insertion point minus 1 is within the range, or something like that.\n\nAlternatively, here's how to approach it: For each i in list_a, check if there exists a j in list_b such that j >= i -k and j <= i +k. Since list_b is sorted, we can perform a binary search to find the first j >= (i -k), and then check if that j is <= (i +k). If yes, then there's a valid j. If that first j is larger than (i +k), then no.\n\nBut what if there's a j in list_b that is less than (i -k)? Well, no, because then j would be less than i -k, and i -j >k. So no. So the only thing to check is whether in the list_b there's an element in [i -k, i +k]. So the binary search approach would work here.\n\nSo for example, using the bisect module in Python. Let's think of list_b is sorted. For each i in list_a, the lower bound is i -k, upper bound is i +k. So find the leftmost index in list_b where j >= lower. Then check if that j is <= upper. If yes, then exists. Or, even if there is an element before that (like if the insertion point is 0, then check if list_b[0] is within the range).\n\nSo the plan is:\n\n- Preprocess list_a and list_b. Find all occurrences of a and b in s. Also, note that the occurrences must be such that the substring starts at i and ends at i + len(a) - 1, which must be within s. So for example, for a of length m, all indices i where 0 <= i <= len(s) - m. Similarly for b.\n\nFirst, how to find all occurrences of a in s. For each possible starting index i, check if s starts with a at i. Similarly for b.\n\nBut how to do that efficiently? Well, for each i in 0 to len(s) - len(a) + 1, check if s[i:i + len(a)] == a. But for len(s) up to 1e5 and a up to 10 characters, this would be O(n * m), which for n=1e5 and m=10 is O(1e6), acceptable.\n\nSo let's proceed.\n\nOnce we have list_a and list_b, sorted (since we generate them in order), then for each element in list_a, check if there exists a j in list_b within [i -k, i +k]. \n\nNow, the steps:\n\nStep 1: Generate list_a and list_b.\n\nFor generating list_a:\n\nLoop from i = 0 to len(s) - len(a). For each i, check if s starting at i equals a. If yes, add to list_a.\n\nSame for list_b.\n\nBut how to do that efficiently? Well, for small a and b (up to 10 chars), as per constraints, even for s up to 1e5, it's manageable.\n\nOnce list_a and list_b are generated, then for each i in list_a, check if there exists a j in list_b such that |i - j| <=k.\n\nBut for each i in list_a, how to check this efficiently?\n\nLet me think about the binary search approach again. Since list_b is sorted, for each i, we can find the possible j's. For example, we can find the position where i -k would be inserted into list_b. Let's say that position is pos. Then, if pos is 0, check list_b[0]. If pos is equal to len(list_b), check the last element. Otherwise, check if list_b[pos] is <= i +k. Or check if list_b[pos -1] is >= i -k?\n\nWait, perhaps a better way: For each i, find the smallest j in list_b that is >= (i -k). Then check if that j is <= (i +k). Also, check if there is any element before that position (if pos >0) which may be in the range. Wait, no. Because if j is >= i -k, then the first element in list_b that is >= i -k is the first candidate. If that element is <= i +k, then yes. Otherwise, no. Because all elements after that are larger than i -k, but maybe larger than i +k. And elements before are less than i -k. So no.\n\nWait, j has to satisfy j >= i -k and j <= i +k. So if the first element >= i -k is larger than i +k, then there is no such j. Otherwise, there is at least one. For example, if the first element >= i -k is x, and x <= i +k, then yes. Because x is a j in the required range.\n\nYes. So the approach is:\n\nFor each i in list_a:\n\n- Compute low = i -k\n\n- Compute high = i +k\n\n- Find the first element in list_b that is >= low. Let's call it pos.\n\n- If there is such an element (pos < len(list_b)), and list_b[pos] <= high, then return True.\n\n- Also, need to check if there are elements before pos? No, because those elements are < low (i -k). So j is less than i -k. Then |i -j| = i -j >k. Not allowed.\n\nWait, no. Suppose j is in [i, i +k]. Then j is >= i -k. So if the first element >= low is within high, then yes. Otherwise, no.\n\nSo this ",
    "### Question:\nYou are given an array of integers nums of length n.\nThe cost of an array is the value of its first element. For example, the cost of [1,2,3] is 1 while the cost of [3,4,1] is 3.\nYou need to divide nums into 3 disjoint contiguous subarrays.\nReturn the minimum possible sum of the cost of these subarrays.\n \nExample 1:\n\nInput: nums = [1,2,3,12]\nOutput: 6\nExplanation: The best possible way to form 3 subarrays is: [1], [2], and [3,12] at a total cost of 1 + 2 + 3 = 6.\nThe other possible ways to form 3 subarrays are:\n- [1], [2,3], and [12] at a total cost of 1 + 2 + 12 = 15.\n- [1,2], [3], and [12] at a total cost of 1 + 3 + 12 = 16.\n\nExample 2:\n\nInput: nums = [5,4,3]\nOutput: 12\nExplanation: The best possible way to form 3 subarrays is: [5], [4], and [3] at a total cost of 5 + 4 + 3 = 12.\nIt can be shown that 12 is the minimum cost achievable.\n\nExample 3:\n\nInput: nums = [10,3,1,1]\nOutput: 12\nExplanation: The best possible way to form 3 subarrays is: [10,3], [1], and [1] at a total cost of 10 + 1 + 1 = 12.\nIt can be shown that 12 is the minimum cost achievable.\n\n \nConstraints:\n\n3 <= n <= 50\n1 <= nums[i] <= 50\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def minimumCost(self, nums: List[int]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to divide an array into three contiguous subarrays and find the minimum possible sum of their first elements. Let me think about how to approach this.\n\nFirst, the problem says the subarrays must be contiguous and disjoint. So, for example, if the array is [1,2,3,12], the possible ways to split into three are:\n\n- [1], [2], [3,12]\n- [1], [2,3], [12]\n- [1,2], [3], [12]\n\nSo the splits are determined by choosing two indices to split. The first split is after some elements, the second split is after a later element. For example, in the first case, splits after index 0 and 1 (assuming 0-based). Wait, let me think about how to model the splits.\n\nLet's say the array is of length n. We need to make two cuts: the first cut is after index i, and the second is after index j, where 1 <= i < j < n-1. Because each subarray must have at least one element. Since there are three subarrays, the first split divides the array into the first subarray (from 0 to i inclusive?), then the second subarray starts from i+1. Then the second split is after j, so the third subarray starts at j+1. Wait, no. Let me think again.\n\nIf we have the array divided into three parts:\n\nsub1 is from index 0 to a (inclusive)\n\nsub2 is from a+1 to b (inclusive)\n\nsub3 is from b+1 to end (inclusive)\n\nSo for this to work, a must be at least 0 (so sub1 has at least 1 element), b must be such that there's at least one element for sub2 and sub3. So for example, the earliest a can be is 0 (so sub1 is [nums[0]], then sub2 starts at 1. Then the earliest j (for b) is 1 (so sub2 is [nums[1]], sub3 starts at 2). The latest a can be is such that there are at least two elements left for sub2 and sub3. For example, if the array has n elements, then a must be at most n-3, because after a, there's n-1 - a elements left. To have at least two, a can be up to n-3. Similarly, j (the index for the second split) would be up to n-2, but depends on a.\n\nBut for all possible splits, we need to consider all possible a and j (b) where 0 <= a <= n-3, and a < j <= n-2. Wait, no. Once a is chosen, the second split (b) must be at least a+1 (so that sub2 has at least one element between a+1 and b, which would be just a+1 to a+1, so b can be a+1. Then sub3 starts at b+1, which would be a+2. For sub3 to have at least one element, b+1 must be <= n-1. So j (b) can be up to n-1 - 1 = n-2, as you said.\n\nSo for each possible a and j where 0 <= a < j <= n-2, we can split the array into three parts. But for the cost, we need the first element of each subarray.\n\nThe first subarray starts at 0, so its first element is nums[0].\n\nThe second subarray starts at a+1. Because it starts at index a+1, so the first element is nums[a+1].\n\nThe third subarray starts at j+1. So first element is nums[j+1].\n\nSo the total cost for a given a and j is nums[0] + nums[a+1] + nums[j+1].\n\nBut we need to make sure that j is greater than a. Because the first split is a, then the second split j must be after a. So for example, a can be from 0 to n-3 (so that there are enough elements left for sub2 and sub3). Then, for each a, j can range from a+1 to n-2. Because j is the split between sub2 and sub3. So sub2 starts at a+1 and ends at j, which requires that j is >= a+1, and sub3 starts at j+1 and ends at n-1. So for each a, j can be from a+1 up to n-2.\n\nTherefore, the problem reduces to finding the minimum value of nums[0] + nums[a+1] + nums[j+1], where 0 <= a < j <= n-2.\n\nWait, that seems correct. Because:\n\nThe first subarray is from 0 to a (inclusive). Then the second subarray is from a+1 to j (inclusive). Then third is j+1 to end.\n\nSo for the second subarray's first element is nums[a+1], third's is nums[j+1]. And the first subarray's first element is nums[0]. So the total cost is the sum of these three.\n\nSo the problem is now to find the minimum sum of these three values, considering all valid a and j.\n\nSo how can we compute this?\n\nWell, given that n is up to 50, we can iterate through all possible a and j values.\n\nFor example, for each a in 0 to n-3:\n\n   for each j in a+1 to n-2:\n\n       calculate the sum and track the minimum.\n\nYes. Because for a in 0 to n-3 (since a can't be higher than n-3, because then there's not enough elements left for sub2 and sub3. For example, if a is n-3, then j can be from a+1 = n-2 + 1? Wait wait. Let me take n=5. For example, n=5, indexes 0-4. Suppose a is 0 (so first subarray is [0..0], then sub2 starts at 1. j can be up to n-2=3. Because sub3 starts at j+1, which can be up to 4 (when j=3). So for a=0, j can be 1,2,3. For example, j=3 is allowed. So for a=0, j can be up to 3 (n-2).\n\nSo for general n, a ranges from 0 to n-3? Because if a is n-3, then the sub2 starts at (n-3)+1 = n-2. Then sub2 can have 1 element (if j is n-2). Then sub3 starts at (n-2)+1 = n-1. So that works.\n\nSo for a, possible values are 0 <= a <= n-3.\n\nFor each a, j can be from a+1 up to n-2.\n\nSo the code can iterate through all possible a and j as described, compute the sum nums[0] + nums[a+1] + nums[j+1], and track the minimum.\n\nBut wait, for example in example 1:\n\nnums = [1,2,3,12]. n=4.\n\nn-3 is 1. So a can be 0 or 1? Wait 0 to n-3 inclusive. For n=4, n-3=1 (since 4-3=1). So a can be 0 or 1?\n\nBut for a to be up to n-3, because after that, j can be a+1 up to n-2. Let's see.\n\nFor example, if a is n-3, then a+1 = n-2. Then j must be >= a+1 = n-2. But j's upper limit is n-2, so j can only be n-2. So for each a, the possible j's are from a+1 to n-2. So for a given a, the earliest possible j is a+1. Because j must be larger than a.\n\nSo for the code:\n\nInitialize min_total to a large value.\n\nLoop over a from 0 to n-3:\n\n   loop over j from a+1 to n-2:\n\n       calculate current_sum = nums[0] + nums[a+1] + nums[j+1]\n\n       if current_sum < min_total, update min_total.\n\nFinally return min_total.\n\nYes. So for example, in example 1, n=4. So n-2 is 2. So j can be from a+1 up to 2.\n\nLet's see:\n\nPossible a values are 0 and 1 (since n-3=1).\n\nFor a=0:\n\nj can be from 1 to 2.\n\nFor j=1: sum is nums[0] + nums[1] + nums[2] → 1+2+3=6.\n\nFor j=2: sum is 1 + 2 + 12 → 15. So for a=0, the minimum is 6.\n\nFor a=1:\n\nj can be from 2 to 2 (since a+1=2). j=2.\n\nsum is nums[0] + nums[2] + nums[3] → 1 + 3 + 12=16. So the overall min is 6. Which matches the example.\n\nYes. So the approach seems correct.\n\nAnother example, example 3:\n\nInput: nums = [10,3,1,1]. n=4. Same as example 1, but the numbers are different.\n\nn=4, so possible a is 0 and 1.\n\nFor a=0 (so first subarray ends at 0, second starts at 1. Then j can be 1 or 2.\n\nWait j can be up to n-2=2.\n\nSo j ranges:\n\nFor a=0, j can be 1 and 2.\n\nFor j=1: sum is 10 (nums[0]) + nums[1] (3) + nums[2] (1) → 14.\n\nFor j=2: 10 + 3 + nums[3] (1) → 14. So both are 14.\n\nFor a=1: j can be 2.\n\nsum is 10 + nums[2] (1) + nums",
    "### Question:\nYou are given an array nums of length n and a positive integer k.\nA subarray of nums is called good if the absolute difference between its first and last element is exactly k, in other words, the subarray nums[i..j] is good if |nums[i] - nums[j]| == k.\nReturn the maximum sum of a good subarray of nums. If there are no good subarrays, return 0.\n \nExample 1:\n\nInput: nums = [1,2,3,4,5,6], k = 1\nOutput: 11\nExplanation: The absolute difference between the first and last element must be 1 for a good subarray. All the good subarrays are: [1,2], [2,3], [3,4], [4,5], and [5,6]. The maximum subarray sum is 11 for the subarray [5,6].\n\nExample 2:\n\nInput: nums = [-1,3,2,4,5], k = 3\nOutput: 11\nExplanation: The absolute difference between the first and last element must be 3 for a good subarray. All the good subarrays are: [-1,3,2], and [2,4,5]. The maximum subarray sum is 11 for the subarray [2,4,5].\n\nExample 3:\n\nInput: nums = [-1,-2,-3,-4], k = 2\nOutput: -6\nExplanation: The absolute difference between the first and last element must be 2 for a good subarray. All the good subarrays are: [-1,-2,-3], and [-2,-3,-4]. The maximum subarray sum is -6 for the subarray [-1,-2,-3].\n\n \nConstraints:\n\n2 <= nums.length <= 10^5\n-10^9 <= nums[i] <= 10^9\n1 <= k <= 10^9\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def maximumSubarraySum(self, nums: List[int], k: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the maximum sum of a good subarray. Let's think about what a good subarray is. A good subarray is one where the absolute difference between the first and last element is exactly k. And I need to find the maximum sum among all such subarrays. If there are none, return 0.\n\nHmm. So, first, how do I approach this? Let me think about the examples given. For example 1, when the array is [1,2,3,4,5,6] and k=1, the good subarrays are all consecutive pairs. The maximum sum is 5+6=11. But since the problem says subarray, which I think here refers to contiguous elements. Wait, yes, because a subarray in array problems is typically a contiguous sequence. So for each subarray nums[i..j], it must be contiguous.\n\nSo the challenge is to find all possible subarrays where |nums[i] - nums[j]| =k, and compute their sum, then find the maximum sum.\n\nBut the problem is that if I naively check all possible subarrays, that would be O(n^2) time, which is not feasible for n up to 1e5. So for n=1e5, O(n^2) would be 1e10 operations, which is way too slow. So I need a smarter approach.\n\nSo how can I find the pairs (i,j) where j >=i + something (since the subarray has to have at least two elements?), but |nums[i] - nums[j]| =k. Wait, for a subarray from i to j, the first element is nums[i], last is nums[j], so the indices are i and j. So for each j, I need to find some i's such that |nums[i] - nums[j]| =k. But how?\n\nLet me think: For each j, we can look for previous elements (i's) where nums[i] = nums[j] ±k. And then among those i's, find the ones where j -i is at least 1 (so the subarray has at least two elements). For each such i, the subarray is from i to j. Then, compute the sum of elements from i to j, and track the maximum.\n\nBut how to efficiently compute the sum from i to j? Well, we can precompute prefix sums. Let's say prefix[j+1] is sum of elements from 0 to j. Then sum from i to j is prefix[j+1] - prefix[i]. So if I have that, then for each j, if I have possible i's that satisfy the condition, I can calculate the sum quickly.\n\nBut the problem is, for each j, how do I efficiently find all the i's that are before j and satisfy nums[i] = nums[j] ±k. Also, for each such i, I need to know the sum of the subarray i to j. But then, for each occurrence of a value in nums, I need to track the indices where they occur, perhaps?\n\nAlternatively, for each j, we can check if there are any nums[i] = nums[j] +k or nums[j] -k. Let me think. Let's consider two cases:\n\nCase 1: nums[i] - nums[j] =k → nums[i] = nums[j] +k. Then the absolute difference is k. But since we take absolute value, another case is nums[i] = nums[j] -k. So for each j, I need to check for both possibilities.\n\nSo for each j, I can look for all previous i's (i < j) where nums[i] = nums[j] +k or nums[j] -k. For each such i, the subarray is from i to j. The sum of this subarray is prefix_sum[j+1] - prefix_sum[i]. So for each j, if I can find the i's that meet the criteria, and compute this difference, then track the maximum.\n\nBut how to efficiently find the i's for each j? Well, for each j, if I have a hash map that maps the value of nums[i] to a list of indices i where this occurs. But building such a map for all possible values would take O(n) space if we consider each occurrence. But for large numbers, that's okay since Python can handle it.\n\nBut even better, perhaps for each j, I can check for the two possible values (nums[j]+k and nums[j]-k) and see if there are any i's with those values before j. And for those i's, I need to find the one that gives the maximum (prefix_sum[j+1] - prefix_sum[i]). To maximize this value, since the prefix_sum[j+1] is fixed for a given j, we need to minimize prefix_sum[i]. Because subtracting a smaller prefix_sum[i] gives a larger value.\n\nWait, yes! For each possible i that meets the criteria (i.e., nums[i] is either nums[j]+k or nums[j]-k), the maximum sum for that j would be prefix_sum[j+1] minus the minimum prefix_sum[i] among all such i's. Because that would give the maximum possible difference.\n\nSo the key insight here is that for each j, if we can track for certain values (like nums[i] values) the minimum prefix sum up to that index. So, for example, for each value v, if we can keep track of the minimum prefix_sum[i] where nums[i] =v. Then, for j, when we check if there are any i's with v equal to nums[j]+k or nums[j]-k, we can look up the minimum prefix sum for those values, and compute the sum for the subarray.\n\nBut how to maintain this? Let's think. We can have a dictionary that maps a value v to the minimum prefix_sum[i] encountered so far. Wait, but maybe there are multiple occurrences of v. For each occurrence of v, we can track the minimum prefix_sum up to that occurrence? Or perhaps for each occurrence of v, we track the minimum prefix_sum[i] for all i's up to current j?\n\nWait, no. Because for a given j, when considering possible i's (i < j), the prefix_sum[i] is the sum up to index i-1? Let me clarify:\n\nLet me define prefix_sum as the array where prefix_sum[0] = 0, and prefix_sum[i] = nums[0] + ... + nums[i-1]. So for index i (the original index in nums is 0-based), the sum from nums[i] to nums[j] (inclusive) is prefix_sum[j+1] - prefix_sum[i].\n\nSo for each j, when looking for i's (i ranges from 0 to j-1), if nums[i] is a certain value, then the sum from i to j is prefix_sum[j+1] - prefix_sum[i]. To maximize this sum, we need to minimize prefix_sum[i].\n\nTherefore, for each value v, if we track the minimum prefix_sum[i] where nums[i] =v, then for each j, when looking for v = nums[j] +k or v= nums[j]-k, we can check if that v exists in the map, and take the minimum prefix_sum[i], which would give the maximum possible sum for that j.\n\nBut how to maintain this map as we iterate through the array?\n\nLet me think. For each index j from 0 to n-1:\n\n- For j=0, there's nothing to do yet. But starting from j=0, we can track for each value v (nums[i] for i up to j) the minimum prefix_sum[i].\n\nWait, but for the next j's, when they look back, they need to look at i's that are before them. So for j, we can process all previous i's (i < j) by maintaining a dictionary that for each value v, stores the minimum prefix_sum[i] where nums[i] =v. But this dictionary is built incrementally. For example, when processing j, before calculating the possible candidates for good subarrays, we first check for the two possible values (nums[j]+k and nums[j]-k) in the map, then then compute the sum. Then, we add the current nums[j] and prefix_sum[j] (since for future j's, this j will be considered as i,",
    "### Question:\nYou are given a 0-indexed string word and an integer k.\nAt every second, you must perform the following operations:\n\nRemove the first k characters of word.\nAdd any k characters to the end of word.\n\nNote that you do not necessarily need to add the same characters that you removed. However, you must perform both operations at every second.\nReturn the minimum time greater than zero required for word to revert to its initial state.\n \nExample 1:\n\nInput: word = \"abacaba\", k = 3\nOutput: 2\nExplanation: At the 1st second, we remove characters \"aba\" from the prefix of word, and add characters \"bac\" to the end of word. Thus, word becomes equal to \"cababac\".\nAt the 2nd second, we remove characters \"cab\" from the prefix of word, and add \"aba\" to the end of word. Thus, word becomes equal to \"abacaba\" and reverts to its initial state.\nIt can be shown that 2 seconds is the minimum time greater than zero required for word to revert to its initial state.\n\nExample 2:\n\nInput: word = \"abacaba\", k = 4\nOutput: 1\nExplanation: At the 1st second, we remove characters \"abac\" from the prefix of word, and add characters \"caba\" to the end of word. Thus, word becomes equal to \"abacaba\" and reverts to its initial state.\nIt can be shown that 1 second is the minimum time greater than zero required for word to revert to its initial state.\n\nExample 3:\n\nInput: word = \"abcbabcd\", k = 2\nOutput: 4\nExplanation: At every second, we will remove the first 2 characters of word, and add the same characters to the end of word.\nAfter 4 seconds, word becomes equal to \"abcbabcd\" and reverts to its initial state.\nIt can be shown that 4 seconds is the minimum time greater than zero required for word to revert to its initial state.\n\n \nConstraints:\n\n1 <= word.length <= 50 \n1 <= k <= word.length\nword consists only of lowercase English letters.\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def minimumTimeToInitialState(self, word: str, k: int) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to find the minimum time greater than zero where the word returns to its initial state after performing operations each second. Each operation involves removing the first k characters and adding any k characters to the end. \n\nHmm. So after each second, the word's length remains the same because we remove k and add k. So the length is always the same as the original. Let's think about how the word changes over time.\n\nLet's denote the original word as S. After each step, the first k characters are removed. But then we add some k characters. However, the problem says that we can add any characters. But for the word to return to its original state after t steps, the added characters must somehow complement the removals.\n\nBut how can we model this? Let's think about each step. Suppose after each operation, the first k chars are taken out, and something is added. Let's think about what remains after each step.\n\nFor example, after the first operation, the word becomes (original word with first k chars removed) plus some new k chars. For the word to return to original after t steps, the operations must have effectively rotated the original string in some way. But since the added chars can be anything, perhaps the problem reduces to finding the earliest time t where the original string is possible by considering that after t operations, the remaining part of the original string (after removing t*k chars from the front) must match the original string's suffix. Because whatever is added in each step could be chosen to build up the original string again.\n\nWait, let's think: After t operations, the first t*k characters have been removed from the original string. But since each time we add k characters, the current word is (original string after removing the first t*k characters) concatenated with some other characters. For the word to be equal to the original string, the part that remains after removing t*k characters from the start must be a prefix of the original string. Because the current word is [original_str[t*k:]] + ... something. And this entire thing must equal the original string. \n\nSo for that to be possible, original_str[t*k:] must be a prefix of the original string. Because when you remove t*k characters, the remaining part (original_str[t*k:]) must match the beginning of the original string. Because when you add k characters each time, you could add whatever needed. But to get back the original string, the remaining part must be the prefix. \n\nFor example, in example 1: word is \"abacaba\", length 7. k=3. Let's compute for t steps:\n\nt=1: remove 3*1=3 chars. original_str[3:] is 'aba' (positions 3,4,5,6? Wait original indexes are 0-based. Original length is 7. For t=1, total removed is 3*1=3. So the remaining part is from index 3 to 6: original_str[3:] is \"caba\". Wait no, the original word is \"abacaba\". Let me check:\n\nIndices 0-6. For t=1, after removing first 3 chars (indices 0,1,2: 'aba'), the remaining part is indices 3-6: 'caba'? Wait original is 7 characters. So after removing 3, the remaining is 4 characters. So original_str[3:] is 'caba' (positions 3: 'c', 4:'a', 5:'b', 6:'a'). Then the current word is 'caba' + some added characters. For the entire word to be original, the 'caba' must be a prefix of the original. But the original starts with 'aba...'. So it's not. Therefore, t=1 is not sufficient. For t=2, total removed is 3*2=6. original_str[6:] is the substring starting at index 6, which is 'a'. Then adding 1 character (since k=3? Wait no. Wait for t steps, the number of characters removed is t*k. So for t=2, 2*3=6, which is possible since original length is 7. The remaining part is original_str[6:] which is length 1. But the current word is that part plus added k* t characters. Wait no, the length of the word remains the same as original. Because each operation removes k and adds k. So after t operations, the first part is original_str[t*k : ] and then some added stuff. The length is len(word) = original length. \n\nSo for the current word to be original_str, original_str[t*k : ] must be a prefix of original_str. Because the added part can be anything, but to get back the original string, after the remaining part (original_str[t*k:]) is the prefix, then adding the rest (the part that would be in original_str) would allow it. \n\nSo the condition is that original_str must equal (original_str[t*k:]) followed by (original_str[ (t*k): + len(remaining) ? No. Wait original_str is of length L. original_str[t*k:] is of length L - t*k. For the entire word (after operations) to be original_str, then the part after original_str[t*k:] must be equal to original_str[0 : L - (L - t*k)] ]? Wait maybe not. Let me think again.\n\nAfter t operations, the current word is made up of the original string's suffix starting at position t*k, plus some added characters. For this word to be equal to the original string, the original string must start with that suffix. So:\n\noriginal_str must equal original_str[t*k:] + ... something. But since the added part can be anything, the minimal time is when original_str starts with original_str[t*k:]. Because then, the added part can be the rest of the original string. \n\nFor example, in example 1, after t=2 operations, we have removed 2*3=6 characters. original_str[6:] is 'a'. The original string starts with 'a', so adding the first 1 character (since the total length is 7, original_str[t*k:] is length 7 - 6 = 1. So the current word is 'a' + ... Well, to make it equal to original_str, the added part must be 'b a c a b a'? Wait but how much is added? Each operation adds k characters. For each of the t steps, we add k characters. But the current word after t steps is (original_str starting from t*k) + (some added parts). How do these added parts affect the state?\n\nWait maybe I'm overcomplicating. Let me think differently. Let me think about what must be true for the word to be equal to the original after t steps. \n\nAfter t operations, the first t*k characters have been removed. Then, in each operation, we added some ",
    "### Question:\nYou are given a 0-indexed string array words.\nLet's define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2:\n\nisPrefixAndSuffix(str1, str2) returns true if str1 is both a prefix and a suffix of str2, and false otherwise.\n\nFor example, isPrefixAndSuffix(\"aba\", \"ababa\") is true because \"aba\" is a prefix of \"ababa\" and also a suffix, but isPrefixAndSuffix(\"abc\", \"abcd\") is false.\nReturn an integer denoting the number of index pairs (i, j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.\n \nExample 1:\n\nInput: words = [\"a\",\"aba\",\"ababa\",\"aa\"]\nOutput: 4\nExplanation: In this example, the counted index pairs are:\ni = 0 and j = 1 because isPrefixAndSuffix(\"a\", \"aba\") is true.\ni = 0 and j = 2 because isPrefixAndSuffix(\"a\", \"ababa\") is true.\ni = 0 and j = 3 because isPrefixAndSuffix(\"a\", \"aa\") is true.\ni = 1 and j = 2 because isPrefixAndSuffix(\"aba\", \"ababa\") is true.\nTherefore, the answer is 4.\nExample 2:\n\nInput: words = [\"pa\",\"papa\",\"ma\",\"mama\"]\nOutput: 2\nExplanation: In this example, the counted index pairs are:\ni = 0 and j = 1 because isPrefixAndSuffix(\"pa\", \"papa\") is true.\ni = 2 and j = 3 because isPrefixAndSuffix(\"ma\", \"mama\") is true.\nTherefore, the answer is 2.  \nExample 3:\n\nInput: words = [\"abab\",\"ab\"]\nOutput: 0\nExplanation: In this example, the only valid index pair is i = 0 and j = 1, and isPrefixAndSuffix(\"abab\", \"ab\") is false.\nTherefore, the answer is 0.\n \nConstraints:\n\n1 <= words.length <= 50\n1 <= words[i].length <= 10\nwords[i] consists only of lowercase English letters.\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def countPrefixSuffixPairs(self, words: List[str]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the number of index pairs (i, j) such that i < j and the word at i is both a prefix and suffix of the word at j. Let me think about how to approach this.\n\nFirst, let's understand the function isPrefixAndSuffix. For str1 to be both a prefix and suffix of str2, str2 must start with str1 and end with str1. So, for example, if str1 is \"aba\" and str2 is \"ababa\", the first three characters are \"aba\" (prefix) and the last three are also \"aba\" (suffix). So the function returns true.\n\nSo for each pair (i,j) where i < j, I need to check if words[i] is both a prefix and suffix of words[j]. Then count all such pairs.\n\nConstraints are small: words length is up to 50, and each word is up to 10 characters. So checking all possible pairs is feasible. Because for 50 words, the maximum number of pairs is 50*49/2 = 1225, which is manageable.\n\nSo the plan is:\n\n1. Iterate over all possible pairs (i, j) with i < j.\n2. For each pair, check if words[i] is both a prefix and a suffix of words[j].\n3. Count how many times this is true.\n\nNow, how to check if a string is both a prefix and suffix of another.\n\nFor a string s to be a prefix of another string t, t must start with s. Similarly for suffix, t must end with s. So in Python, for a given s and t, we can check:\n\nlen(s) <= len(t), and t.startswith(s) and t.endswith(s).\n\nWait, yes. But how efficient is that? Well, for small strings, it's fine. So for each pair where len(words[i]) <= len(words[j]), we can perform those checks.\n\nSo for each pair (i, j) where i < j:\n\nif len(words[i]) > len(words[j]): skip.\n\nelse:\n\ncheck if words[j].startswith(s) and words[j].endswith(s), where s is words[i].\n\nIf both are true, increment the count.\n\nSo now, the problem becomes coding this in Python.\n\nLet's see the examples.\n\nExample 1:\n\nwords = [\"a\",\"aba\",\"ababa\",\"aa\"]\n\nFor i=0, j=1: \"a\" length is 1, aba is length 3. starts and ends with 'a' → yes.\n\nSimilarly for j=2 and 3.\n\ni=1, j=2: \"aba\" is length 3, \"ababa\" is length 5. Let's see:\n\nstarts with: ababa starts with aba? Yes. Ends with aba? The last three characters are 'aba' (positions 2-4: indices 2,3,4 in 0-based). So yes. So returns true.\n\nSo that would be counted.\n\nSo the code would check all pairs.\n\nSo in code:\n\nInitialize a count to 0.\n\nLoop i from 0 to len(words)-1:\n\n   loop j from i+1 to len(words)-1:\n\n      s = words[i]\n\n      t = words[j]\n\n      if len(s) > len(t): skip.\n\n      else:\n\n          if t.startswith(s) and t.endswith(s):\n\n              count += 1\n\nReturn count.\n\nYes, that seems right.\n\nBut wait, what about cases where s is longer than t? Well, in that case, it's impossible for s to be a prefix or suffix, so we can skip checking.\n\nSo that's handled by the len(s) <= len(t) check.\n\nSo that's the approach.\n\nNow, implement this in Python.\n\nWait, the code for the class would be:\n\nclass Solution:\n    def countPrefixSuffixPairs(self, words: List[str]) -> int:\n        count = 0\n        n = len(words)\n        for i in range(n):\n            for j in range(i+1, n):\n                s = words[i]\n                t = words[j]\n                if len(s) > len(t):\n                    continue\n                if t.startswith(s) and t.endswith(s):\n                    count +=1\n        return count\n\nWait, but in example 1, the answer is 4, let's test this code.\n\nIn example 1:\n\nCheck all pairs:\n\ni=0, j=1 → yes. count 1.\n\ni=0, j=2 → yes. count 2.\n\ni=0, j=3 → t is \"aa\". starts and ends with \"a\"? Yes. So count 3.\n\ni=1, j=2: s is \"aba\", t is \"ababa\". starts with yes, ends with yes. count 4.\n\ni=1, j=3: s is \"aba\", t is \"aa\". len(s)=3, len(t)=2 → skip.\n\ni=0, j=3 done.\n\nOther pairs:\n\ni=1, j=3: no.\n\ni=2, j=3: len(aba...?) ababa is length 5, t is aa (length 2). len(s) is 5 > 2 → skip.\n\ni=0, j=1,2,3 → 3 counts.\n\nThen i=1, j=2: counts. So total 4, which matches the example.\n\nAnother example:\n\nExample 2:\n\nwords = [\"pa\",\"papa\",\"ma\",\"mama\"]\n\nPossible pairs:\n\ni=0, j=1 → pa vs papa. len 2 <= 4. papa starts and ends with pa? Starts yes. Ends with pa? The end is 'pa' (positions 2 and 3). Yes. So count 1.\n\ni=0, j=2: pa vs ma. len is 2 vs 2. starts with? No. ends with? No. So no.\n\ni=0, j=3: len 2 <=4. starts with pa? Yes. ends with pa? The last two letters of mama are 'ma' → no. So no.\n\nSo no count here.\n\ni=1, j=2: papa vs ma. len 4 vs 2 → skip.\n\ni=1, j=3: len 4 vs 4? papa and mama. starts with? No. ends with? No.\n\ni=2, j=3: ma vs mama. len 2 <=4. starts with? yes. ends with? yes. So count increases by 1. Total 2.\n\nWhich matches the example.\n\nSo the code works here.\n\nThird example:\n\n[\"abab\",\"ab\"]\n\nCheck pairs (i=0, j=1). s is abab, t is ab. len(s) is 4, len(t) is 2. So skip. So no count. Which matches the example.\n\nAnother test case.\n\nWhat about a case where s is both prefix and suffix, but in between?\n\nLike words = [\"a\", \"aba\"], which would have the pair (0,1). The code counts it. Let's see, \"aba\" starts with \"a\" and ends with \"a\". So yes. So count is 1.\n\nSo code works here.\n\nWhat about if the word is exactly the same as s? For example, words = [\"a\", \"a\", \"a\"]. The pairs (0,1), (0,2), (1,2). For each pair, since the other word is same, so starts and ends with. So total 3 counts. The code would return 3.\n\nYes.\n\nSo this approach seems correct.\n\nNow, code-wise, in Python, the startswith and endswith methods are available for strings. So the code is straightforward.\n\nSo the code is as above. Let me write it properly.\n\nThe starter code is given, and we have to implement it.\n\nWait, the code uses List[str], so need to import List from typing. But in LeetCode's environment, that's handled.\n\nSo the code should work.\n\nNow check for possible edge cases.\n\nEdge case 1: All words are same length. For example, words = [\"a\", \"a\"], then i < j. Check if \"a\" is prefix and suffix of \"a\". Well, since len(s) is same as len(t). So, t.startswith(s) is True, and t.endswith(s) is True. So yes, count 1.\n\nAnother case: words[i] is longer than words[j], so skipped.\n\nAnother case: words[j] is exactly s + something + s. Like s is \"ab\", t is \"abxab\". Wait, t starts with \"ab\" and ends with \"ab\"? The ends with check: t ends with s. So for t of length 5, s is length 2. The last two characters are 'ab'? Yes. So code would count it.\n\nYes.\n\nSo the code seems correct.\n\nSo the answer is the code as written.\n</think>To solve this problem, we need to count the number of index p",
    "### Question:\nYou are given a 0-indexed string array words having length n and containing 0-indexed strings.\nYou are allowed to perform the following operation any number of times (including zero):\n\nChoose integers i, j, x, and y such that 0 <= i, j < n, 0 <= x < words[i].length, 0 <= y < words[j].length, and swap the characters words[i][x] and words[j][y].\n\nReturn an integer denoting the maximum number of palindromes words can contain, after performing some operations.\nNote: i and j may be equal during an operation.\n \nExample 1:\n\nInput: words = [\"abbb\",\"ba\",\"aa\"]\nOutput: 3\nExplanation: In this example, one way to get the maximum number of palindromes is:\nChoose i = 0, j = 1, x = 0, y = 0, so we swap words[0][0] and words[1][0]. words becomes [\"bbbb\",\"aa\",\"aa\"].\nAll strings in words are now palindromes.\nHence, the maximum number of palindromes achievable is 3.\nExample 2:\n\nInput: words = [\"abc\",\"ab\"]\nOutput: 2\nExplanation: In this example, one way to get the maximum number of palindromes is: \nChoose i = 0, j = 1, x = 1, y = 0, so we swap words[0][1] and words[1][0]. words becomes [\"aac\",\"bb\"].\nChoose i = 0, j = 0, x = 1, y = 2, so we swap words[0][1] and words[0][2]. words becomes [\"aca\",\"bb\"].\nBoth strings are now palindromes.\nHence, the maximum number of palindromes achievable is 2.\n\nExample 3:\n\nInput: words = [\"cd\",\"ef\",\"a\"]\nOutput: 1\nExplanation: In this example, there is no need to perform any operation.\nThere is one palindrome in words \"a\".\nIt can be shown that it is not possible to get more than one palindrome after any number of operations.\nHence, the answer is 1.\n \nConstraints:\n\n1 <= words.length <= 1000\n1 <= words[i].length <= 100\nwords[i] consists only of lowercase English letters.\n\n### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.\n```python\nclass Solution:\n    def maxPalindromesAfterOperations(self, words: List[str]) -> int:\n        \n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to find the maximum number of palindromes possible in the words array after performing any number of swaps. Let me think about how to approach this.\n\nFirst, I remember that a palindrome can be formed if we have even counts of characters, except possibly one odd count (for the middle character in an odd-length palindrome). Since we can swap any characters between any positions in any words, the key is to figure out how to optimally distribute the characters to form as many palindromes as possible.\n\nWait, the swaps allow us to rearrange all characters freely. Because when you can swap any characters between any positions, effectively, each word can be treated as a collection of characters. The order of characters within a word matters only insofar as forming a palindrome. But since we can rearrange the characters, the problem becomes similar to: given the total count of all characters, how can we distribute them across the words to maximize the number of palindromes.\n\nSo the steps I need to take:\n\n1. Count all the characters across all words. Let's say we have a frequency map for all 26 letters.\n\n2. Then, figure out how to allocate these frequencies to form as many palindromes as possible, considering each word's length.\n\nBut how to model this? Let's think about the requirements for each palindrome.\n\nEach palindrome can be made up of even pairs and at most one single character (for odd length). For multiple palindromes, each will require their own set of characters.\n\nThe main idea is to maximize the number of palindromes. Let's think about the total number of pairs we can form from all characters.\n\nFor example, for each character, the number of pairs it contributes is floor(count / 2). Summing over all characters gives total possible pairs. Then, to form palindromes, each palindrome of length L needs (floor(L/2)) pairs. Additionally, we can have at most one single character left in each palindrome.\n\nBut since we can arrange the pairs and singles across the words, how do we maximize the number of palindromes?\n\nLet me think of this as follows:\n\nThe maximum number of palindromes is determined by two parts:\n\n- The total number of pairs available. Each palindrome requires a certain number of pairs based on its length. For example, a word of length m (after any rearrangement) will contribute floor(m/2) required pairs. The sum of all required pairs for all words must be <= total available pairs.\n\nBut since we can choose how to assign lengths (since we can split the characters between words as needed), perhaps we need to minimize the total required pairs by choosing the words with smaller lengths first. Because shorter words require fewer pairs.\n\nSo the approach would be:\n\n- Decide for which words to make them palindromes. To maximize the count, we should prioritize making as many as possible, even if they are shorter.\n\nSo here's a plan:\n\n1. Calculate the total count of each character across all words.\n\n2. For the pairs: calculate the sum over all characters of (count // 2), which is the total available pairs.\n\n3. For the single characters: sum over (count % 2) for all characters. But since each swap can allow us to collect even counts except for one, but when forming palindromes, each palindrome can take one single character. However, the total number of single characters available is not directly important here because we can redistribute the pairs and singles. Wait, perhaps the main constraint is the pairs and the number of palindromes.\n\nLet me think again. For each palindrome, the number of pairs needed depends on the length of the word. For example, a word of length l can be a palindrome by using k pairs (k = floor(l / 2)), and possibly one single character. So for each word that we want to turn into a palindrome, we need to allocate k pairs, and then one single character (if the length is odd).\n\nSo the problem becomes:\n\nWe need to select a subset of words, and for each, assign a certain number of pairs and possibly a single character. The total pairs used must not exceed the total available pairs. Also, the number of single characters needed is equal to the number of palindromes that have an odd length. And the maximum possible number of such single characters is the sum of (1 for each palindrome with odd length). However, the total number of available single characters can be calculated as (total characters) - 2*(total pairs). Wait, no. Because initially, the total number of single characters (after taking out pairs) is sum(count % 2 for all characters). But since we can swap any characters, perhaps the single characters can be any characters. However, when forming the palindromes, the single characters can be chosen from any available, but each palindrome can take one. However, the actual number of available single characters is not fixed once the pairs are allocated. Because after using pairs, perhaps we can take some more characters from pairs to break them into singles, but that would decrease the total pairs.\n\nWait, no. Because the pairs are formed as much as possible. Once you have the total pairs, you can also have some single characters left. For example, if you have 3 a's, that's one pair and one single. But if you decide to not use that single, then you can have more pairs. But once you have the total pairs, the number of possible single characters is equal to (total_characters) - 2*total_pairs. Which is the same as sum(count % 2 for all characters) mod 2? No. Wait, sum(count % 2) is the total number of leftover characters after forming as many pairs as possible. So that sum is equal to total_characters - 2 * total_pairs. So for example, if you have 3 a's, sum(count%2) is 1. Total pairs is 1. So total_characters (3) - 2 * total_pairs (2) = 1. Which equals the sum.\n\nSo, once the total_pairs is fixed, the number of available single characters is equal to total_char - 2 * total_pairs. But since each single character can be used as the center of a palindrome, the number of palindromes we can have is limited by the number of palindromes with odd lengths. Each such palindrome needs exactly one single character.\n",
    "### Question:\nTakahashi and Aoki played N games.\r\nYou are given a string S of length N, representing the results of these games.\r\nTakahashi won the i-th game if the i-th character of S is T, and Aoki won that game if it is A.\nThe overall winner between Takahashi and Aoki is the one who won more games than the other.\r\nIf they had the same number of wins, the overall winner is the one who reached that number of wins first.\r\nFind the overall winner: Takahashi or Aoki.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nS\n\nOutput\n\nIf the overall winner is Takahashi, print T; if it is Aoki, print A.\n\nConstraints\n\n\n- 1\\leq N \\leq 100\n- N is an integer.\n- S is a string of length N consisting of T and A.\n\nSample Input 1\n\n5\r\nTTAAT\n\nSample Output 1\n\nT\r\n\nTakahashi won three games, and Aoki won two.\r\nThus, the overall winner is Takahashi, who won more games.\n\nSample Input 2\n\n6\r\nATTATA\n\nSample Output 2\n\nT\r\n\nBoth Takahashi and Aoki won three games.\r\nTakahashi reached three wins in the fifth game, and Aoki in the sixth game.\r\nThus, the overall winner is Takahashi, who reached three wins first.\n\nSample Input 3\n\n1\r\nA\n\nSample Output 3\n\nA\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to solve this problem where I have to determine the overall winner between Takahashi and Aoki based on the game results. Let's read the problem statement carefully again.\n\nThe problem says that they played N games. The string S represents the results. Each character is T or A. The winner is the one with more wins. If they have the same number, then whoever reached that count first wins.\n\nSo first, I need to count the number of T's and A's. If one is larger, that's the winner. If equal, check when each reached that count first.\n\nHmm. For example, in sample input 2, both have 3 wins. But Takahashi (T) reached 3 wins first in game 5, while Aoki (A) reached 3 in game 6. So the answer is T.\n\nSo how do I track when each reached their count?\n\nLet me think about the approach:\n\nFirst, count total T and A. If they are different, the higher one wins. But if they are same, then during the iteration of the string, I need to track when each reached half of the total (since if total is even, then for example 3 each in 6 games, so half is 3). Wait, but for N games, if there's a tie, each has N/2 wins? Or not necessarily? Because for example, in sample input 2, N is 6, but T reached 3 wins at game 5, not at game 6.\n\nSo, when counts are tied, I need to track during the games, which player first reached the count of (total/2) wins. Wait, no. Because the total is the same. For example, suppose total T and A are both 3. So during processing, when each game is checked, I track the count of T and A. The moment when either count reaches the target (which is 3 in this case), whichever player reached it first.\n\nWait, but for example, in the sample input 2, the counts after each game are:\n\nLet me list the sample input 2: 6, ATTATA.\n\nThe string is A T T A T A.\n\nLet's process each character step by step, keeping track of T and A counts.\n\nGame 1: A → A=1, T=0 → current counts.\n\nGame 2: T → A=1, T=1.\n\nGame 3: T → T=2, A=1.\n\nGame4: A → T=2, A=2.\n\nGame5: T → T=3, A=2. So here, T reaches 3 first. At this point, since the total counts are 3 each, but T reached 3 first. So the answer is T.\n\nSo during the iteration, I need to track the counts as I go. Once either count reaches (total/2) for T or A? Wait, no. Because the total for each is same only when the overall counts are same. So first, after counting the total T and A, if they are same, then during processing, I need to track when each reaches (total T) which is same as (total A) = same number. So for example, in this case, total T is 3, total A is 3. So during processing, whenever either T or A reaches 3 first, that's the winner.\n\nSo the steps for the code:\n\n1. Read N and S.\n\n2. Count the total T and A. Let's say count_T and count_A.\n\n3. If count_T > count_A → print T.\n\n   Else if count_A > count_T → print A.\n\n4. Else (count_T == count_A). Then, during processing the string step by step (from left to right), track the current T and A counts. For each step, check if either count_T_current or count_A_current reaches the target (which is count_T, since they are same). But wait, since the total is count_T = count_A, so for example, if total is 3 each. So during processing, when the current count for T reaches 3, that's when the total is achieved. But also, the same for A. However, since the total is same, whichever of the two reaches their total first is the winner.\n\nSo during iteration, for each character in S:\n\n- increment current T or A count.\n\n- if at any step, current_T equals the total_T (which is same as total_A), then T reaches it first, so winner is T.\n\nSame for if current_A equals total_A, then A is winner.\n\nBut since the total is same, once either reaches the total, that's the earliest possible.\n\nWait, for example, in sample input 2, the T count reaches 3 at step 5. At that moment, the A count is 2. So how do we know that it's T who reaches the total first?\n\nBecause the total for T is 3. Once current_T reaches 3, that's the target. Because the total is 3. So during processing, once any of the counts reaches the total (which is same for both), then that's the winner.\n\nSo in this case, during processing, when T reaches 3, which is the total, then that's the first occurrence. So we can immediately return T as winner.\n\nBut how do I compute this? Let's see.\n\nOnce we know that the total T and A are same (let's call it total), then during iteration, for each step, we track current_T and current_A. For each step, if current_T reaches total before current_A does, then T is winner. Similarly for A.\n\nSo for the sample input 2, during processing:\n\nAt each step, until step 5, current_T is 3, current_A is 2. So since current_T reaches the total (3) first, T is the winner.\n\nBut what if during processing, both current_T and current_A reach the total at the same step?\n\nLike, suppose the total is 2 each. And the string is \"TTAA\". Then, after processing the first two T's, current_T is 2. current_A is 0. But then after processing the third and fourth characters (A's), current_A becomes 2. But in this case, T reached their total earlier.\n\nBut what if a scenario where both reach their totals at the same time? Like impossible? Because for each step, we process one game. So for example, if the string is \"TATA...\".\n\nWait, let's take N=2, S=\"TA\". Then total_T is 1, total_A is 1. During processing:\n\nAfter first character (T), current_T=1, current_A=0. So T reaches their total first. So answer is T.\n\nBut in N=2, S=\"AT\", then current_A reaches 1 first, so answer is A.\n\nBut if N=4, S is \"TTAA\", then during step 2, T is 2, A is 0. So T reaches total (2) first. So answer T.\n\nBut if S is \"TATA\", then step 1: T=1, A=0 → T reaches 1. Since total is 2 each. So no. Wait, no. For N=4, total_T is 2 and total_A is 2. So during processing:\n\nstep 1: T=1, A=0.\n\nstep 2: T=1, A=1.\n\nstep 3: T=2, A=1. → current_T reaches 2. So T is first. So answer is T.\n\nBut if the string is \"TAT A\", like \"TATA\" (N=4). Let's see:\n\nProcessing step by step:\n\nAfter 1st char: T=1, A=0.\n\nAfter 2nd: T=1, A=1.\n\nAfter 3rd: T=2, A=1. → T reaches total, so winner is T.\n\nSo even though the A will reach 2 in step 4, but T already reached their total first.\n\nSo in this case, the code needs to check, during processing step by step, which of the counts reaches their total first.\n\nSo for the case where counts are tied, the code needs to process each character, track current T and A counts, and check if either has reached the total (which is same for both). The first to reach it is the winner.\n\nSo the plan is:\n\nWhen count_T == count_A:\n\n   Initialize current_t = 0, current_a = 0.\n\n   For each character in S:\n\n      increment current_t or current_a.\n\n      after incrementing, check if current_t equals total_t (which is same as total / 2, since count_t + count_a = N",
    "### Question:\nWe have a grid with H rows and W columns.\r\nLet (i,j) denote the square at the i-th row from the top and j-th column from the left.\r\nEach square in the grid is one of the following: the start square, the goal square, an empty square, a wall square, and a candy square.\r\n(i,j) is represented by a character A_{i,j}, and is the start square if A_{i,j}= S, the goal square if A_{i,j}= G, an empty square if A_{i,j}= ., a wall square if A_{i,j}= #, and a candy square if A_{i,j}= o.\r\nHere, it is guaranteed that there are exactly one start, exactly one goal, and at most 18 candy squares.\nTakahashi is now at the start square.\r\nHe can repeat moving to a vertically or horizontally adjacent non-wall square.\r\nHe wants to reach the goal square in at most T moves.\r\nDetermine whether it is possible.\r\nIf it is possible, find the maximum number of candy squares he can visit on the way to the goal square, where he must finish.\r\nEach candy square counts only once, even if it is visited multiple times.\n\nInput\n\nThe input is given from Standard Input in the following format:\nH W T\r\nA_{1,1}A_{1,2}\\dots A_{1,W}\r\n\\vdots\r\nA_{H,1}A_{H,2}\\dots A_{H,W}\n\nOutput\n\nIf it is impossible to reach the goal square in at most T moves, print -1.\r\nOtherwise, print the maximum number of candy squares that can be visited on the way to the goal square, where Takahashi must finish.\n\nConstraints\n\n\n- 1\\leq H,W \\leq 300\n- 1 \\leq T \\leq 2\\times 10^6\n- H, W, and T are integers.\n- A_{i,j} is one of S, G, ., #, and o.\n- Exactly one pair (i,j) satisfies A_{i,j}= S.\n- Exactly one pair (i,j) satisfies A_{i,j}= G.\n- At most 18 pairs (i,j) satisfy A_{i,j}= o.\n\nSample Input 1\n\n3 3 5\r\nS.G\r\no#o\r\n.#.\n\nSample Output 1\n\n1\r\n\nIf he makes four moves as (1,1) \\rightarrow (1,2) \\rightarrow (1,3) \\rightarrow (2,3) \\rightarrow (1,3), he can visit one candy square and finish at the goal square.\r\nHe cannot make five or fewer moves to visit two candy squares and finish at the goal square, so the answer is 1.\nNote that making five moves as (1,1) \\rightarrow (2,1) \\rightarrow (1,1) \\rightarrow (1,2) \\rightarrow (1,3) \\rightarrow (2,3) to visit two candy squares is invalid since he would not finish at the goal square.\n\nSample Input 2\n\n3 3 1\r\nS.G\r\n.#o\r\no#.\n\nSample Output 2\n\n-1\r\n\nHe cannot reach the goal square in one or fewer moves.\n\nSample Input 3\n\n5 10 2000000\r\nS.o..ooo..\r\n..o..o.o..\r\n..o..ooo..\r\n..o..o.o..\r\n..o..ooo.G\n\nSample Output 3\n\n18\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's try to figure out how to solve this problem. So, the problem is about finding the maximum number of candies Takahashi can collect while reaching the goal within T moves. The grid has walls, empty cells, start and goal, and some candy cells. The key points are that he must reach the goal in at most T steps, and maximize the candies collected along the way.\n\nFirst, I need to think about the constraints. The grid can be up to 300x300, which is 90,000 cells. But there are at most 18 candy squares. That makes me think of a state that includes the number of candies collected so far. Since there are 18, we can represent the state as (position, mask), where mask is a bitmask representing which candies have been collected. The maximum mask would be 2^18 = 262,144 possible masks. But considering that for each cell in the grid and each mask, we need to track the minimum steps required to reach that cell with that mask. \n\nWait, but if we track for each (position, mask) the minimum number of steps to reach that state, then we can later check if from there we can reach the goal in time. But how do we combine the steps?\n\nAlternatively, since the problem requires us to reach the goal, perhaps we can model this as a BFS where each state includes the current position and the set of collected candies (represented by a bitmask). For each such state, we track the minimum number of steps taken to get there. Then, for all states that reach the goal (regardless of the mask), we check if the total steps plus the steps from the goal's position (wait, no, once you reach the goal, you stop. So the state must end at the goal. So during BFS, whenever we reach the goal, we check if the steps taken are <= T. And among all such possibilities, find the maximum number of 1s in the mask.\n\nBut the problem is that for a 300x300 grid and 2^18 masks, the total number of possible states is 90,000 * 262,144 ≈ 23.6 billion. That's way too big for a standard BFS approach, especially since T can be up to 2e6. So this approach isn't feasible.\n\nHmm. So perhaps there's a smarter way. Let's think differently. Since there are only 18 candy squares, maybe we can model the problem as visiting some subset of them in a certain order, and find the optimal path that collects as many as possible while also getting to the goal within T steps.\n\nAnother idea: Let's precompute the shortest distances between all pairs of candy squares (and start and goal). Then, the problem becomes similar to the Traveling Salesman Problem (TSP), where we want to find the order of visiting candies that maximizes the count while ensuring that the total path length plus the distance from start to first candy, between candies, and from last candy to goal is <= T. But since we can choose any subset of candies (up to 18), and the maximum is desired, perhaps we can use dynamic programming for TSP.\n\nBut first, we need to compute the shortest paths between all relevant points. The relevant points are the start (S), the goal (G), and each of the candy squares (o's). Let's say there are K candy squares. Then, there are K+2 points. For each of these points, we need to compute their positions, and then find the minimal distance between any two of them, avoiding walls. Also, note that during the path, Takahashi can collect candies in any order, but must reach G in T steps.\n\nSo the plan is:\n\n1. Find all the positions of S, G, and all the o's. Let's call them nodes. Let's assign each node an index. For example, S is 0, G is K+1, and the o's are numbered from 1 to K.\n\n2. For each pair of nodes, compute the minimal distance between them. If there's no path, then that pair is not possible. However, since the problem requires reaching G, we need that all nodes involved in the path must have a path to G.\n\nOnce we have these distances, we can model the problem as a TSP-like problem where we want to collect as many nodes as possible (the o's) and then reach G, with total distance <= T.\n\nThe maximum number of candies is K, so we can try to find the maximum k such that there exists a path that collects k candies and has total distance <= T.\n\nBut how do we model this?\n\nLet me think. Let's say we have K candy squares. Let's denote the start as S, the goal as G, and each candy as o_0, o_1, ..., o_{K-1}.\n\nWe need to find a path that starts at S, possibly visits some subset of the o's in some order, then ends at G, such that the total steps taken is <= T. The maximum number of o's in the subset is our answer.\n\nThe total steps would be the sum of the minimal distances between consecutive points in the path. For example, if the path is S -> o_i -> o_j -> ... -> o_k -> G, then the total steps is d(S, o_i) + d(o_i, o_j) + ... + d(o_k, G). And this sum must be <= T.\n\nSo, the problem reduces to finding the largest m such that there exists a permutation of m distinct o's, and the sum of the distances from S to first o, between o's, and from last o to G is <= T. Also, if there's a direct path from S to G that's shorter than any such path with candies, that's also considered.\n\nSo, how can we compute this efficiently? Since K is up to 18, a DP approach for TSP is feasible.\n\nThe standard TSP DP is O(K^2 * 2^K), which for K=18 is 18^2 * 2^18 ≈ 486 million operations. Which is manageable if optimized properly.\n\nSo here's the plan:\n\n- Preprocess all the positions of S, G, and o's. Let's say there are K o's. Let's have K+2 points (S, G, o_0, ..., o_{K-1} }).\n\n- For each pair of these points, compute the minimal",
    "### Question:\nOn a two-dimensional plane, Takahashi is initially at point (0, 0), and his initial health is H.  M items to recover health are placed on the plane; the i-th of them is placed at (x_i,y_i).\nTakahashi will make N moves.  The i-th move is as follows.\n\n- \nLet (x,y) be his current coordinates.  He consumes a health of 1 to move to the following point, depending on S_i, the i-th character of S:\n\n- (x+1,y) if S_i is R;\n- (x-1,y) if S_i is L;\n- (x,y+1) if S_i is U;\n- (x,y-1) if S_i is D.\n\n\n- \nIf Takahashi's health has become negative, he collapses and stops moving.  Otherwise, if an item is placed at the point he has moved to, and his health is strictly less than K, then he consumes the item there to make his health K.\n\n\nDetermine if Takahashi can complete the N moves without being stunned.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN M H K\nS\nx_1 y_1\n\\vdots\nx_M y_M\n\nOutput\n\nPrint Yes if he can complete the N moves without being stunned; print No otherwise.\n\nConstraints\n\n\n- 1\\leq N,M,H,K\\leq 2\\times 10^5\n- S is a string of length N consisting of R, L, U, and D.\n- |x_i|,|y_i| \\leq 2\\times 10^5\n- (x_i, y_i) are pairwise distinct.\n- All values in the input are integers, except for S.\n\nSample Input 1\n\n4 2 3 1\nRUDL\n-1 -1\n1 0\n\nSample Output 1\n\nYes\n\nInitially, Takahashi's health is 3.  We describe the moves below.\n\n- \n1-st move: S_i is R, so he moves to point (1,0).  His health reduces to 2.  Although an item is placed at point (1,0), he do not consume it because his health is no less than K=1.\n\n- \n2-nd move: S_i is U, so he moves to point (1,1).  His health reduces to 1.\n\n- \n3-rd move: S_i is D, so he moves to point (1,0).  His health reduces to 0.  An item is placed at point (1,0), and his health is less than K=1, so he consumes the item to make his health 1.\n\n- \n4-th move: S_i is L, so he moves to point (0,0).  His health reduces to 0.\n\n\nThus, he can make the 4 moves without collapsing, so Yes should be printed.  Note that the health may reach 0.\n\nSample Input 2\n\n5 2 1 5\nLDRLD\n0 0\n-1 -1\n\nSample Output 2\n\nNo\n\nInitially, Takahashi's health is 1.  We describe the moves below.\n\n- \n1-st move: S_i is L, so he moves to point (-1,0).  His health reduces to 0.\n\n- \n2-nd move: S_i is D, so he moves to point (-1,-1).  His health reduces to -1.  Now that the health is -1, he collapses and stops moving.\n\n\nThus, he will be stunned, so No should be printed.\nNote that although there is an item at his initial point (0,0), he does not consume it before the 1-st move, because items are only consumed after a move.\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's tackle this problem. Hmm, what's the problem statement again? Oh right, Takahashi starts at (0,0) with health H. He makes N moves, each move costing 1 health. If his health drops below zero, he collapses. Also, there are M items on the plane. Whenever he moves to a position with an item and his current health is strictly less than K, he uses the item to restore his health to K. We need to determine if he can complete all N moves without collapsing.\n\nSo the main steps are to simulate his movement, track his health, and check for the conditions where he can use the items. But given the constraints, especially with N and M up to 2e5, we need an efficient way to check if he's on an item position after each move.\n\nWait, the key thing here is that after each move, we need to check if the current position has an item. And if so, whether to use it. But since using an item requires that his current health is less than K. Also, once an item is used, it's consumed and can't be used again. Oh right, because the items are placed at different positions and once used, they are removed. Wait, the problem says \"if an item is placed at the point he has moved to...\" So does that mean that each item is present only once? Like, if he steps on the same position again after the item has been used, there's no item anymore. But since all (x_i, y_i) are pairwise distinct, each item is unique. So, for example, if there are multiple items on the same position (but according to the input, they are distinct points), no. So each position can have at most one item.\n\nTherefore, during the simulation, after each move, we need to check if the current position has an item. If yes, and current health is less than K, then we use it. And once used, that item is removed from the available items.\n\nThe problem is how to efficiently check for each position whether there's an item. Because for each of the N steps, after moving to a new position, we need to check if that position is in the set of items. For M up to 2e5, using a hash set (like a Python set) would allow O(1) lookups. So, the plan is:\n\n- Store all the item positions in a set. Like, a set of tuples. But in Python, tuples are hashable and can be in a set.\n\nSo the approach is:\n\n1. Read all item positions into a set.\n\n2. Initialize current position (x,y) as (0,0), health as H.\n\n3. For each step in the move string S:\n\n   a. Move according to the direction, subtract 1 from health.\n\n   b. Check if health is negative. If yes, output No.\n\n   c. If not, check if current position is in the item set. If yes, and current health is less than K, then consume the item: remove it from the set (so it's not available again), and set health to K.\n\nSo, that's the simulation plan. The key is that checking for the item positions is O(1) per step, which is manageable for N up to 2e5.\n\nBut wait, step 3a: moving according to direction. For example, initial position is (0,0). Then after first move, depends on S[0].\n\nNow, for the data structures: the set for items would be a Python set containing tuples. For example, for sample input 1, the set would have (-1,-1) and (1,0). When he moves to (1,0) in first step, check if that tuple is in the set. If yes, and health after move is 2 (H=3-1=2, which is >= K=1, so no consumption. Then in step 3, after moving to (1,0) again (after step 3), his health is 0 (H=1-1=0, which is less than K=1). So he uses the item, health becomes 1, and the item at (1,0) is removed from the set.\n\nThis seems correct.\n\nNow, let's think about the sample input 2. Let's see:\n\nSample Input 2:\n\n5 2 1 5\n\nLDRLD\n\n0 0\n\n-1 -1\n\nWait, but M is 2. So there are two items. Let me check the input format. The input for items is M lines, each with x_i and y_i. So in sample 2, there are two items: (0,0) and (-1,-1). Let's see the simulation.\n\nInitial position (0,0), health 1.\n\nFirst move: L → direction is left, so new position (-1, 0). Health becomes 0. Then check if ( -1, 0 ) is in the item set. Are there any items there? No. The items are (0,0) and (-1,-1). So no item here. Then next step.\n\nSecond move: D → down → (-1, -1). Health becomes -1. Now health is negative → collapse. Output is No. Which matches the sample.\n\nSo the simulation works here.\n\nSo the approach seems correct.\n\nNow, implementing this in Python.\n\nBut for Python, the steps are:\n\nRead N, M, H, K.\n\nRead S as a string.\n\nRead M lines of x_i, y_i, and store them in a set. Let's call it items_set.\n\nThen simulate each step in S:\n\nInitialize x, y as 0, 0.\n\ncurrent_health = H.\n\nfor each character in S:\n\n    move the x, y according to direction.\n\n    current_health -= 1.\n\n    if current_health < 0: return No.\n\n    check if (x, y) is in items_set:\n\n        if yes, and current_health < K:\n\n            current_health = K\n\n            remove (x, y) from items_set.\n\nSo, that's the plan.\n\nBut what about the order? For example, after moving to a position, check if there is an item there. Even if there is, but the current_health is exactly K, then we don't use it. Only when it's strictly less than K.\n\nNow, for the data structures, the set is initially built with all the item positions. And once an item is used, it's removed from the set. So subsequent steps won't trigger again for that position.\n\nNow, considering that M can be up to 2e5, and for each of the N steps, we do O(1) operations, the total",
    "### Question:\nYour computer has a keyboard with three keys: 'a' key, Shift key, and Caps Lock key.  The Caps Lock key has a light on it.\nInitially, the light on the Caps Lock key is off, and the screen shows an empty string.\nYou can do the following three actions any number of times in any order:\n\n- Spend X milliseconds to press only the 'a' key.  If the light on the Caps Lock key is off, a is appended to the string on the screen; if it is on, A is.\n- Spend Y milliseconds to press the 'a' key and Shift key simultaneously.  If the light on the Caps Lock key is off, A is appended to the string on the screen; if it is on, a is.\n- Spend Z milliseconds to press the Caps Lock key.  If the light on the Caps Lock key is off, it turns on; if it is on, it turns off.\n\nGiven a string S consisting of A and a, determine at least how many milliseconds you need to spend to make the string shown on the screen equal to S.\n\nInput\n\nThe input is given from Standard Input in the following format:\nX Y Z\nS\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 1 \\leq X,Y,Z \\leq 10^9\n- X, Y, and Z are integers.\n- 1 \\leq |S| \\leq 3 \\times 10^5\n- S is a string consisting of A and a.\n\nSample Input 1\n\n1 3 3\nAAaA\n\nSample Output 1\n\n9\n\nThe following sequence of actions makes the string on the screen equal to AAaA in 9 milliseconds, which is the shortest possible.\n\n- Spend Z(=3) milliseconds to press the CapsLock key.  The light on the Caps Lock key turns on.\n- Spend X(=1) milliseconds to press the 'a' key.  A is appended to the string on the screen.\n- Spend X(=1) milliseconds to press the 'a' key.  A is appended to the string on the screen.\n- Spend Y(=3) milliseconds to press the Shift key and 'a' key simultaneously.  a is appended to the string on the screen.\n- Spend X(=1) milliseconds to press the 'a' key.  A is appended to the string on the screen.\n\nSample Input 2\n\n1 1 100\naAaAaA\n\nSample Output 2\n\n6\n\nSample Input 3\n\n1 2 4\naaAaAaaAAAAaAaaAaAAaaaAAAAA\n\nSample Output 3\n\n40\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to find the minimum time required to type a given string S using the three keys: 'a', Shift, and Caps Lock. The initial state is Caps Lock off and an empty string. \n\nHmm, the problem involves dynamic programming, probably. Because each step depends on the current state—like whether Caps Lock is on or off. Also, maybe the actions taken before affect how we type the next characters optimally.\n\nLet me think about the possible states. The main state here is the Caps Lock status (on or off). For each position in the string, I need to track what the Caps Lock state is, and what's the minimum time to reach that state after typing up to that character.\n\nSo dynamic programming with state tracking. Let's model it as a DP table where dp[i][state] represents the minimum time to type the first i characters with the Caps Lock in 'state' (0 for off, 1 for on). Here, 'state' can be 0 or 1.\n\nFor each position i (starting from 0 up to the length of S), we consider both possibilities for the Caps Lock state. Then, for each possible state, we decide how to type the next character, considering all possible actions that could lead to the next state.\n\nLet me think about how to transition between states. For each step, the next character can be either 'a' or 'A'. Depending on the current Caps Lock state and the desired output, we can choose between different actions.\n\nLet me take an example. Suppose the current Caps Lock state is off. The next character needs to be 'A'. How can I type that?\n\nOption 1: Press Shift + 'a' (cost Y). This appends an uppercase A (since when Caps is off, Shift makes it uppercase). The Caps state remains off.\n\nOption 2: Toggle Caps Lock (cost Z), then press 'a' (cost X). Now, the Caps is on. But after that, the next character is typed. However, this would change the state. So if the next character also requires uppercase, maybe it's better to have Caps on. But for this single character, the cost would be Z + X. Compare this to Y. Which is better? Depends on whether there are more characters to type that would benefit from Caps being on.\n\nBut since the DP tracks the state, we need to consider all possibilities. For example, when considering the next step, we can either stay in the same state (by using Shift) or switch state (by toggling Caps Lock and then typing, which would change the state).\n\nSo for each step, when moving from i to i+1, considering the current Caps state, we can have transitions based on different actions.\n\nLet me formalize the transitions. Let's say for the i-th character (starting from 0?), and for each possible current state (0 or 1), we look at possible ways to type the next character and compute the next state and accumulate the cost.\n\nLet me think of the DP transitions. For each i-th character in S, and for each possible state (Caps on or off), we can consider:\n\nCase 1: Current Caps state is s (s is 0 or 1). The next character is c (a or A).\n\nWe can choose to type this character in three different ways (but considering what's possible):\n\nOption A: Type with current state. For example, if the current state is such that pressing 'a' gives the desired case. Or pressing 'a' with Shift. Or maybe that's not an option. Let's think.\n\nThe desired character is c. Let me think for each possible current state s, what actions can produce c.\n\nFor example, suppose current state is 0 (Caps off):\n\n- Pressing 'a' (cost X) appends 'a'. So if the desired character is 'a', this is good.\n- Pressing Shift+'a' (cost Y) appends 'A'. So if desired is 'A', this is good.\n- Or toggle Caps Lock (cost Z) to state 1, then press 'a' (cost X). So total Z+X. But after this action, the Caps state is 1. So this would be considered as a transition from state 0 to state 1, and adding the cost.\n\nSimilarly, if current state is 1 (Caps on):\n\n- Pressing 'a' gives 'A' (so desired 'A' is good with cost X).\n- Pressing Shift+'a' gives 'a' (desired 'a' with cost Y).\n- Toggle Caps Lock (cost Z) to state 0, then press 'a' (cost X) for total Z+X. But then the state becomes 0.\n\nBut when considering transitions for the DP, perhaps for each step, we can choose between two possibilities: either type the current character without changing the state (using either the 'a' key or shift+'a'), or change the state (by toggling and then typing, which may involve some cost, but also changing the state for future characters).\n\nBut how do these choices affect the DP transitions?\n\nFor each step, when adding the i-th character, the DP[i][new_state] will be updated based on DP[i-1][previous_state] plus the cost of the chosen action.\n\nBut since each character must be added, for each position i, we need to process one character. So for each position, we have to decide how to type that character, and possibly change the state.\n\nLet me formalize:\n\nLet's define dp[i][s] as the minimum time to type the first i characters of S, ending with Caps Lock state s (s is 0 or 1).\n\nInitially, dp[0][0] = 0 (no characters typed, Caps off), and dp[0][1] = Z (if we toggled once). Wait, no. Initially, before typing any characters, the Caps Lock is off. So dp[0][0] = 0, and dp[0][1] = Z (if we toggle Caps Lock once). But maybe there's a possibility that before typing the first character, we could have toggled multiple times. However, that's probably not optimal. For example, toggling twice would cost 2Z, but end up back in the original state. Since Z is positive, it's better to not do that. Therefore, for the initial state, dp[0][0] = 0, and dp[0][1] = Z (cost to toggle once).\n\nBut during the DP transitions, we have to consider that toggling the Caps Lock changes the state. Each toggle costs Z, and flips the state.\n\nSo for each i-th step (i from 0 to n-1?), for each state (0 or 1), we can consider how to type the (i+1)th character.\n\nLet me think of the transitions for dp[i+1][new_state], based on the current state and the action taken.\n\nLet me take for example, the current state is s (0 ",
    "### Question:\nThere is a rectangular cake with some strawberries on the xy-plane. The cake occupies the rectangular area \\lbrace (x, y) : 0 \\leq x \\leq W, 0 \\leq y \\leq H \\rbrace.\nThere are N strawberries on the cake, and the coordinates of the i-th strawberry are (p_i, q_i) for i = 1, 2, \\ldots, N. No two strawberries have the same coordinates.\nTakahashi will cut the cake into several pieces with a knife, as follows.\n\n- First, cut the cake along A different lines parallel to the y-axis: lines x = a_1, x = a_2, \\ldots, x = a_A.\n- Next, cut the cake along B different lines parallel to the x-axis: lines y = b_1, y = b_2, \\ldots, y = b_B.\n\nAs a result, the cake will be divided into (A+1)(B+1) rectangular pieces. Takahashi will choose just one of these pieces to eat. Print the minimum and maximum possible numbers of strawberries on the chosen piece.\nHere, it is guaranteed that there are no strawberries along the edges of the final pieces. For a more formal description, refer to the constraints below.\n\nInput\n\nThe input is given from Standard Input in the following format:\nW H\nN\np_1 q_1\np_2 q_2\n\\vdots\np_N q_N\nA\na_1 a_2 \\ldots a_A\nB\nb_1 b_2 \\ldots b_B\n\nOutput\n\nPrint the minimum possible number of strawberries m and the maximum possible number M on the chosen piece in the following format, separated by a space.\nm M\n\nConstraints\n\n\n- 3 \\leq W, H \\leq 10^9\n- 1 \\leq N \\leq 2 \\times 10^5\n- 0 \\lt p_i \\lt W\n- 0 \\lt q_i \\lt H\n- i \\neq j \\implies (p_i, q_i) \\neq (p_j, q_j)\n- 1 \\leq A, B \\leq 2 \\times 10^5\n- 0 \\lt a_1 \\lt a_2 \\lt \\cdots \\lt a_A \\lt W\n- 0 \\lt b_1 \\lt b_2 \\lt \\cdots \\lt b_B \\lt H\n- p_i \\not \\in \\lbrace a_1, a_2, \\ldots, a_A \\rbrace\n- q_i \\not \\in \\lbrace b_1, b_2, \\ldots, b_B \\rbrace\n- All input values are integers.\n\nSample Input 1\n\n7 6\n5\n6 1\n3 1\n4 2\n1 5\n6 2\n2\n2 5\n2\n3 4\n\nSample Output 1\n\n0 2\n\nThere are nine pieces in total: six with zero strawberries, one with one strawberry, and two with two strawberries. Therefore, when choosing just one of these pieces to eat, the minimum possible number of strawberries on the chosen piece is 0, and the maximum possible number is 2.\n\nSample Input 2\n\n4 4\n4\n1 1\n3 1\n3 3\n1 3\n1\n2\n1\n2\n\nSample Output 2\n\n1 1\n\nEach piece has one strawberry on it.\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to solve this problem where I have to find the minimum and maximum number of strawberries in a particular piece of cake after some cuts. Let me think about how to approach this.\n\nThe cake is divided into vertical and horizontal slices. The vertical cuts are lines x = a_1, a_2,...a_A. So these divide the width into A+1 columns. Similarly, the horizontal cuts are B different lines, dividing the height into B+1 rows. Each piece is a rectangle defined by consecutive cuts in x and y directions.\n\nEach strawberry lies in exactly one of these pieces. Because the problem says that no strawberry is on the edge, so for any strawberry, its x coordinate isn't in any a_i, and same for y and b_j.\n\nThe task is to compute the minimum and maximum count of strawberries in all the possible pieces.\n\nFirst, how to determine which piece each strawberry is in?\n\nFor each strawberry (p_i, q_i), we need to find:\n\n- Which vertical slice it's in. For example, the vertical positions are divided into regions like [0, a_1), [a_1, a_2), ..., [a_A, W]. Wait, no, since the cuts are at a_1, etc., the regions are between 0 to a_1, a_1 to a_2, etc. So for a strawberry's x-coordinate p_i, we need to find the index k such that a_{k} < p_i < a_{k+1}. But since the a's are sorted, we can use binary search here. Similarly for the y-coordinate and the b array.\n\nOnce we can find for each strawberry which vertical and horizontal strip it's in, then we can group the strawberries by their (vertical region, horizontal region) pairs. Each such pair corresponds to a piece of cake. The count of strawberries in each piece is the number of strawberries assigned to that (region x, region y) pair.\n\nSo the problem reduces to:\n\nFor all possible (i,j) where i ranges from 0 to A, j ranges from 0 to B, count how many strawberries are in that cell. Then, since each piece is one of these cells, the min and max among these counts are our answers.\n\nBut how to compute for each cell?\n\nThe idea is to process each strawberry and assign it to a cell, then count per cell.\n\nBut how to find the cell coordinates?\n\nFor example, for the vertical cuts a = [a_1, a_2,...a_A], sorted in increasing order. For a p_i, find the largest a_k that is less than p_i. The index k will determine which vertical slice it's in. For example, if a is [2,5], then for p_i=3, it's in the first slice between a_0 (which is 0) and a_1 (2)? No, wait, a_1 is 2 and a_2 is 5. The regions are 0-2, 2-5, 5-7 (assuming W=7 as in sample 1). Wait, for W=7 and cuts at 2 and 5, the three vertical regions are 0 < x <2, 2 <x <5, and 5 <x <7. So for p_i=3, it's in the second vertical slice (index 1?), because between 2 and 5. Wait, how to index them? Let's think: the vertical slices are between 0 and a_1, then between a_1 and a_2, etc. So the number of vertical slices is A+1. For each p_i, the vertical position is in one of these intervals. To find which interval, we can perform a binary search on the a array. For example, if the a array is sorted, find the index where a_k is the largest value less than p_i. The index (plus 1?) would determine the vertical slice. For example, if all a's are in a list, then using bisect.bisect_left or bisect.bisect_right?\n\nLet me think. Let's say the a array is sorted. For a given p_i, the correct interval is the one where a[k] < p_i < a[k+1]. For example, for the list [2,5], and p_i=3. The bisect.bisect_left on 3 would give index 1 (since 3 is bigger than 2, and inserted before 5). Wait, no. Wait, bisect_left returns the insertion point. For example, in the list [2,5], searching for 3 would return index 1, because 3 is between 2 and 5. Wait no, wait bisect_left finds the first position where the element can be inserted. For example, in a list sorted in increasing order. For example, for 3 in [2,5], the insertion point would be 1. So the number of elements less than 3 is 1. So the slice is between a[0] (2) and a[1] (5). So the index of the vertical slice is 1 (if we start counting from 0). Because there are 1 elements in the a array that are less than p_i. So for the vertical slice index is the number of vertical cuts that are to the left of p_i. So the vertical slice is determined by that count.\n\nSo for each strawberry, we can find the vertical index v_i (number of a's less than p_i) and horizontal index h_i (number of b's less than q_i). Then, each (v_i, h_i) pair represents a unique cell, and we can count how many strawberries are in each cell.\n\nOnce we have these counts, the maximum is the maximum count, and the minimum is the minimum count. However, there's a catch: some cells may have zero strawberries. For example, in sample input 1, the output has 0 as the minimum. How do we handle that?\n\nBecause not all cells necessarily contain a strawberry. So, the minimum could be zero if there are some empty cells. However, how do we know how many cells are empty?\n\nFor example, suppose there are (A+1) * (B+1) cells. If all cells have at least zero, but some are empty, then the minimum is zero. But how can we confirm that there is at least one empty cell?\n\nSo, during processing, when we count the strawberries per cell, the cells that have strawberries are accounted for. But the minimum could be zero if there's at least one cell without any strawberries.\n\nBut how do we check if there exists a cell with zero strawberries?\n\nWell, the total number of cells is (A+1)*(B+1). Let's say the total number of cells is C. If N (number of strawberries) is equal to C, then every cell has exactly one strawberry. But if N is less than C, then there are some cells with zero. So, the minimum possible would be zero in that case.\n\nWait, but maybe some cells have more than one strawberry. So ",
    "### Question:\nYou are given an undirected graph G with N vertices and M edges.\r\nFor i = 1, 2, \\ldots, M, the i-th edge is an undirected edge connecting vertices u_i and v_i.\nA graph with N vertices is called good if the following condition holds for all i = 1, 2, \\ldots, K:\n\n- there is no path connecting vertices x_i and y_i in G.\n\nThe given graph G is good.\nYou are given Q independent questions. Answer all of them.\r\nFor i = 1, 2, \\ldots, Q, the i-th question is as follows.\n\n- Is the graph G^{(i)} obtained by adding an undirected edge connecting vertices p_i and q_i to the given graph G good?\n\nInput\n\nThe input is given from Standard Input in the following format:\nN M\r\nu_1 v_1\r\nu_2 v_2\r\n\\vdots\r\nu_M v_M\r\nK\r\nx_1 y_1\r\nx_2 y_2\r\n\\vdots\r\nx_K y_K\r\nQ\r\np_1 q_1\r\np_2 q_2\r\n\\vdots\r\np_Q q_Q\n\nOutput\n\nPrint Q lines.\r\nFor i = 1, 2, \\ldots, Q, the i-th line should contain the answer to the i-th question: Yes if the graph G^{(i)} is good, and No otherwise.\n\nConstraints\n\n\n- 2 \\leq N \\leq 2 \\times 10^5\n- 0 \\leq M \\leq 2 \\times10^5\n- 1 \\leq u_i, v_i \\leq N\n- 1 \\leq K \\leq 2 \\times 10^5\n- 1 \\leq x_i, y_i \\leq N\n- x_i \\neq y_i\n- i \\neq j \\implies \\lbrace x_i, y_i \\rbrace \\neq \\lbrace x_j, y_j \\rbrace\n- For all i = 1, 2, \\ldots, K, there is no path connecting vertices x_i and y_i.\n- 1 \\leq Q \\leq 2 \\times 10^5\n- 1 \\leq p_i, q_i \\leq N\n- p_i \\neq q_i\n- All input values are integers.\n\nSample Input 1\n\n6 6\r\n1 2\r\n2 3\r\n2 3\r\n3 1\r\n5 4\r\n5 5\r\n3\r\n1 5\r\n2 6\r\n4 3\r\n4\r\n2 5\r\n2 6\r\n5 6\r\n5 4\n\nSample Output 1\n\nNo\r\nNo\r\nYes\r\nYes\r\n\n\n- For the first question, the graph G^{(1)} is not good because it has a path 1 \\rightarrow 2 \\rightarrow 5 connecting vertices x_1 = 1 and y_1 = 5. Therefore, print No.\n- For the second question, the graph G^{(2)} is not good because it has a path 2 \\rightarrow 6 connecting vertices x_2 = 2 and y_2 = 6. Therefore, print No.\n- For the third question, the graph G^{(3)} is good. Therefore, print Yes.\n- For the fourth question, the graph G^{(4)} is good. Therefore, print Yes.\n\nAs seen in this sample input, note that the given graph G may have self-loops or multi-edges.\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this programming problem. Let's read the problem carefully.\n\nSo, we have an undirected graph G with N vertices and M edges. The graph is considered \"good\" if for all K specified pairs (x_i, y_i), there is no path connecting them. The task is to answer Q queries, each asking whether adding a new edge (p_i, q_i) to G would make the graph no longer good. For each query, we need to output \"Yes\" if the resulting graph is still good, else \"No\".\n\nConstraints are tight: up to 2e5 for N, M, K, Q. So any O(Q*K) approach would be impossible. Need an efficient way.\n\nFirst, the initial graph is good. That means for each forbidden pair (x_i, y_i), they are in different connected components. Adding a new edge may connect some components. The problem is to check, for each query edge (p, q), whether adding it would connect any of the forbidden pairs.\n\nBut how do we check that efficiently?\n\nLet me think. For each query, adding the edge (p, q) could potentially merge the connected components of p and q. If before adding this edge, there exists a forbidden pair (x_i, y_i) such that x_i is in the component of p and y_i is in the component of q (or vice versa), then adding this edge would create a path between x_i and y_i, making the graph not good. Also, if adding the edge doesn't merge any components (because p and q were already connected), then it's possible that no forbidden pair is connected. So, for each query, we need to check if adding (p, q) connects any forbidden pair.\n\nBut how?\n\nAn approach could be:\n\nFor each forbidden pair (x_i, y_i), they are in different components in the original graph. Let's precompute for each forbidden pair the connected component they belong to. For example, for each forbidden pair (x_i, y_i), let's store the root of their component in the DSU (Disjoint Set Union) structure. But since the original graph is fixed, we can compute the DSU once.\n\nNow, when adding a new edge (p, q), which would merge the components of p and q. For the query, we need to check whether there exists any forbidden pair where x_i is in the original component of p and y_i is in the original component of q. Because adding the edge would merge those two components, which could connect some forbidden pairs.\n\nWait, but if in the original DSU, p and q are already in the same component, then adding the edge does nothing. So, in that case, the graph remains good if it was good before. Since the original graph is good, adding an edge that doesn't merge components won't affect anything. Therefore, in this case, the answer is \"Yes\".\n\nSo, the key idea is:\n\nFor each query (p, q):\n\n1. Find the original DSU roots of p and q. Let's call them u and v.\n\nIf u == v: adding the edge does nothing. The graph remains good. Output \"Yes\".\n\nElse: Adding the edge will merge u and v. We need to check if there exists any forbidden pair (x_i, y_i) such that one endpoint is in u's component and the other is in v's component. If such a pair exists, then adding the edge would connect x_i and y_i, making the graph invalid. Hence, answer \"No\". Otherwise, \"Yes\".\n\nSo the problem reduces to, for each query where u and v are different, check if any forbidden pair has one in the component u and the other in component v.\n\nBut how do we check this efficiently?\n\nThe problem is that K can be up to 2e5. For each query, checking all forbidden pairs would be O(K) per query, which is O(2e5 * 2e5) = 4e10 operations. That's way too slow.\n\nWe need a smarter way.\n\nAlternative idea: For each forbidden pair (x_i, y_i), record their DSU root in the original graph. Let's say for each forbidden pair, we have (a, b), where a is the root of x_i and b is the root of y_i (but since DSU roots can vary, perhaps we should make sure that each component is represented by a unique identifier, like the DSU's find function returns). Wait, in DSU, for each node, find gives the root. So for each forbidden pair (x_i, y_i), since they are not connected, their roots are different. Let's pre-process all forbidden pairs as (root_x_i, root_y_i), sorted in a way that, say, the smaller root comes first (to avoid duplicates). For example, for each forbidden pair, we can store (min(a, b), max(a, b)), representing the pair of roots that must not be connected.\n\nThen, for each query (p, q), which connects components u and v (original roots). After merging, any forbidden pair that has one end in u's component and the other in v's component will now be connected.\n\nTherefore, the problem becomes: for the original DSU, when merging u and v, does the set of forbidden pairs include any (u, v) pair?\n\nWait, no. Because the forbidden pairs are between different original components. So, for each forbidden pair (a, b), where a and b are original roots (a != b), if adding the edge u-v would cause a pair (a, b) where a is in u's component and b is in v's component, or vice versa, then adding the edge would connect a and b.\n\nBut how do we represent that?\n\nSo, for each forbidden pair, it's a pair of original roots. Let's collect all these (a, b) pairs. For each query, if the original roots of p and q are u and v (u != v), then we need to check if (u, v) or (v, u) exists in the forbidden pairs. If yes, then adding the edge will connect this forbidden pair. But also, there might be forbidden pairs where one is in u's component and the other in v's component.\n\nWait, no. Because when you merge u and v, any forbidden pair that has one in u's component and another in v's component will now be connected. So, for example, if there's a forbidden pair (a, b) where a is in u's component and b is in v's component, then merging u and v would connect a and b.\n\nSo, the set of forbidden pairs that would be affected by merging u and v is all pairs (a, b) where (a is u and b is v) OR (a is in u's component and b is in v's component) OR vi",
    "### Question:\nYou are given a string S of length N consisting of lowercase English letters.\nWe denote the i-th character of S by S_i.\nPrint the string of length 2N obtained by concatenating S_1,S_1,S_2,S_2,\\dots,S_N, and S_N in this order.\nFor example, if S is beginner, print bbeeggiinnnneerr.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\nS\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- N is an integer such that 1 \\le N \\le 50.\n- S is a string of length N consisting of lowercase English letters.\n\nSample Input 1\n\n8\nbeginner\n\nSample Output 1\n\nbbeeggiinnnneerr\n\nIt is the same as the example described in the problem statement.\n\nSample Input 2\n\n3\naaa\n\nSample Output 2\n\naaaaaa\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to solve this problem where I'm given a string S of length N, and I have to create a new string where each character in the original is repeated twice. Oh right, like for each character in S, I put it twice in the new string. So for example, if the input is 'abc', the output is 'aabbcc'.\n\nHmm, how to approach this. Well, the input comes as N followed by S. Wait, but the length of S should be N. But according to the problem statement, that's guaranteed, so I don't need to check that.\n\nSo the main task is to process each character in S and duplicate it. How can I do that in Python?\n\nLet me think. For each character in the string S, I can take the character and add it twice to a new string. So, maybe loop through each character and build the result step by step.\n\nLike, initialize an empty string. Then for every char c in S, append c twice. Then finally output that.\n\nYes. For example, in Python:\n\nresult = ''\nfor c in S:\n    result += c * 2\nprint(result)\n\nThat should work. Let me test with sample input 2. Input is 'aaa', so each a is duplicated, leading to 6 a's. Which matches the sample output.\n\nAnother sample input is 'beginner' with length 8. Let's see. The first character is 'b', so two 'b's. Then 'e' becomes 'ee', etc. So the sample output starts with 'bb' then 'ee' etc. Which matches.\n\nSo the code would be straightforward. Now, how to read the input?\n\nThe input is given as N on the first line, then S on the next line. So in Python, I can read them as:\n\nn = int(input())\ns = input().strip()\n\nBut wait, do I need to check that the length of s is equal to n? According to the problem statement, S is of length N. So it's given correctly, so no need.\n\nThen process as described.\n\nSo putting it all together:\n\nRead n, read s. Then loop through each character in s, duplicate it, and build the result.\n\nYes.\n\nAnother test case: if input is N=1, S='a', output is 'aa'. Which should be correct.\n\nEdge cases: N=0? No, constraints say N is at least 1. So no problem.\n\nSo the code would be:\n\nRead N, S.\n\nresult = ''\n\nfor c in S:\n    result += c * 2\n\nprint(result)\n\nThat's it. Let me code this in Python.\n\nWait, but in Python, strings are handled easily. Another way could be to use a list to append each duplicated character and then join. But for small N (up to 50), it's not a problem either way.\n\nSo code:\n\nn = int(input())\ns = input().strip()\n\nans = ''\nfor c in s:\n    ans += c * 2\n\nprint(ans)\n\nYes. Let me check with sample input 1. Let's take sample input 1:\n\n8\nbeginner\n\nThe string is 'beginner'. Let's process each character:\n\nb → bb, e→ee, g→gg, i→ii, n→nn, n→nn, e→ee, r→rr. Wait, but the sample output is 'bbeeggiinnnneerr'. Wait, let me check the sample input.\n\nWait the problem says for S=beginner, the output is bbeeggiinnnneerr. Let me see:\n\nOriginal S: b e g i n n e r. Each character is duplicated. So:\n\nb b e e g g i i n n n n e e r r. Wait, but the sample output is written as \"bbeeggiinnnneerr\". Let me count:\n\nbbeeggiinnnneerr. Let's break it down:\n\nbb (from first b), then ee (from e), then gg (from g), then ii (from i), then nn (from first n?), then another nn? Because the original S has two n's. So for each of the two n's in S, we have two n's each. So two n's in original would contribute four n's. So in the sample output, after the i's, there are four n's. Then ee (from the next e) and rr (from r). So the sample output is correct.\n\nWhich would be the case if each character is duplicated. So the code is correct.\n\nTherefore, the code should work.\n</think>```python\nn = int(input())\ns = input().strip()\n\nresult = ''\nfor char in s:\n    result += char * 2\n\nprint(result)\n```",
    "### Question:\nYou are given N strings S_1,S_2,\\ldots,S_N consisting of lowercase English letters.\r\nDetermine if there are distinct integers i and j between 1 and N, inclusive, such that the concatenation of S_i and S_j in this order is a palindrome.\nA string T of length M is a palindrome if and only if the i-th character and the (M+1-i)-th character of T are the same for every 1\\leq i\\leq M.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nS_1\r\nS_2\r\n\\vdots\r\nS_N\n\nOutput\n\nIf there are i and j that satisfy the condition in the problem statement, print Yes; otherwise, print No.\n\nConstraints\n\n\n- 2\\leq N\\leq 100\n- 1\\leq \\lvert S_i\\rvert \\leq 50\n- N is an integer.\n- S_i is a string consisting of lowercase English letters.\n- All S_i are distinct.\n\nSample Input 1\n\n5\r\nab\r\nccef\r\nda\r\na\r\nfe\n\nSample Output 1\n\nYes\r\n\nIf we take (i,j)=(1,4), the concatenation of S_1=ab and S_4=a in this order is aba, which is a palindrome, satisfying the condition.\r\nThus, print Yes.  \nHere, we can also take (i,j)=(5,2), for which the concatenation of S_5=fe and S_2=ccef in this order is feccef, satisfying the condition.\n\nSample Input 2\n\n3\r\na\r\nb\r\naba\n\nSample Output 2\n\nNo\r\n\nNo two distinct strings among S_1, S_2, and S_3 form a palindrome when concatenated.\r\nThus, print No.\r\nNote that the i and j in the statement must be distinct.\n\nSample Input 3\n\n2\r\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\r\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n\nSample Output 3\n\nYes\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. The problem is to determine if there are two distinct indices i and j such that when you concatenate S_i and S_j, the result is a palindrome. Hmm. So I need to check all possible pairs of distinct i and j and see if any of their concatenations form a palindrome.\n\nFirst, let's think about the constraints. N can be up to 100, so the number of possible pairs is 100 * 99 = 9900, which is manageable. So the plan is: iterate through all possible pairs (i,j) where i != j, check if S_i + S_j is a palindrome. If any of them is, output \"Yes\"; otherwise, \"No\".\n\nBut how to check if a string is a palindrome? Well, the standard way is to compare the string with its reverse. Like, if s == s[::-1], then it's a palindrome. That's straightforward.\n\nBut wait, for each pair, concatenating the two strings and then checking might be acceptable given the constraints. Each string can be up to 50 characters, so the maximum length of the concatenated string is 100 characters. Checking 9900 such strings, each up to 100 characters, that's feasible.\n\nSo the steps are:\n\n1. Read N.\n2. Read N strings into a list, say, strings.\n3. For each i from 0 to N-1:\n   a. For each j from 0 to N-1 where j != i:\n      i. Concatenate strings[i] and strings[j] into temp.\n      ii. Check if temp is a palindrome.\n      iii. If yes, return \"Yes\".\n4. After checking all pairs, if none found, return \"No\".\n\nBut wait, the problem says \"distinct integers i and j between 1 and N\", so in code terms, for all i and j where i != j. So in the loops, for each possible pair where i and j are different.\n\nBut in code, how to implement this? For example, in Python, for each i in range(n), then for j in range(n), and if j != i. Since N is small, this is acceptable.\n\nNow, considering that N is up to 100, and each check is O(M) where M is up to 100 (since each string is up to 50 characters), the total time is acceptable.\n\nBut what about the sample input 2? Let's take the sample input 2:\n\n3\na\nb\naba\n\nThe possible pairs are:\n\na + b: ab → not palindrome.\n\na + aba: aaba → check if it's a palindrome. aaba reversed is abaa. Not same.\n\nb + a: ba → no.\n\nb + aba: babaa → reversed is aabab. Not same.\n\naba + a: abaa → reversed is aaba → no.\n\naba + b: abab → reversed is baba → no.\n\nSo output is No.\n\nSample input 1 has pairs like ab + a → aba, which is a palindrome. So the code would find that.\n\nNow, let's think about edge cases. For example, when N=2. Suppose the two strings, when concatenated in both possible orders. For example, if the two strings are \"a\" and \"a\". Wait, but according to the constraints, all S_i are distinct. So in the input, all the strings are different. So no need to worry about same strings. But for N=2, there are two possible pairs: (i=0,j=1) and (i=1,j=0). So both need to be checked.\n\nAnother edge case: one of the strings is empty? No, according to the constraints, each S_i has length at least 1.\n\nAnother thing: when you concatenate S_i and S_j, the total length can be even or odd. But the palindrome check is straightforward.\n\nSo the code plan is solid.\n\nNow, coding steps.\n\nRead N, then N strings. Then check all possible pairs where i != j.\n\nBut in Python, reading N strings can be done via a list. Let's code:\n\nRead N from input.\n\nThen read N lines, stripping any whitespace (probably, the inputs don't have leading/trailing spaces, but better to read with .strip() to be safe).\n\nThen for each i in 0..N-1, and j in 0..N-1, i != j:\n\nBut wait, since for each possible pair (i,j), including (i,j) and (j,i). Both need to be checked. Because, for example, in sample input 1, (1,4) gives \"ab\"+\"a\" = \"aba\", which is a palindrome. But (4,1) would be \"a\"+\"ab\" = \"aab\", which is not a palindrome. So checking only i and j in some order may miss the valid case. Hence, need to check all possible ordered pairs where i != j.\n\nSo the code should loop for all possible ordered pairs (i,j) with i ≠ j.\n\nSo in code:\n\nfor i in range(n):\n    for j in range(n):\n        if i != j:\n            concat = strings[i] + strings[j]\n            if concat == concat[::-1]:\n                print(\"Yes\")\n                exit()\nprint(\"No\")\n\nBut wait, in Python, exiting early can be done with return, but in a script, perhaps using sys.exit(). But for code submission, perhaps better to check each pair, and as soon as one is found, print \"Yes\" and exit. So yes.\n\nBut in code submission, the code must read from stdin and output to stdout. So here's the plan:\n\nRead N, then read N lines, then for each i and j:\n\nBut in code:\n\nimport sys\n\nn = int(sys.stdin.readline())\nstrings = [sys.stdin.readline().strip() for _ in range(n)]\n\nfor i in range(n):\n    for j in range(n):\n        if i != j:\n            s = strings[i] + strings[j]\n            if s == s[::-1]:\n                print(\"Yes\")\n                exit()\n\nprint(\"No\")\n\nWait, but in Python, the exit() function may not be allowed? Or perhaps, using sys.exit(). Oh right. So in code, after importing sys, we can do:\n\nOnce a pair is found, print \"Yes\" and exit. Otherwise, after checking all pairs, print \"No\".\n\nThis code should handle all cases.\n\nLet's test sample input 1:\n\nSample input 1:\n\n5\n\nab\n\nccef\n\nda\n\na\n\nfe\n\nThe code would check all pairs. Let's see for (i=0, j=3): ab + a → aba → yes. So it would print Yes and exit.\n\nSimilarly, (i=4, j=1): fe + ccef → feccef. Let's check: fecc ef → reversed would be feccfe? Wait, let me see. The string is 'feccef'. Let's reverse it: 'feccef' reversed is 'feccef'? Wait, no. Let me check:\n\nOriginal string: f e c c e f. Reverse would be f e c c e f. Oh right, so it's a palindrome. So yes.\n\nIn code, when i=4 (string 'fe') and j=1 (string 'ccef'), the concatenation is 'feccef', which reversed is same. So code will find this as well. But since the code checks in order, once any pair is found, it outputs yes.\n\nSo sample input 1 returns yes.\n\nSample input 2: 3 strings a, b, aba. Let's see. All possible pairs:\n\nCheck (0,1): a + b → ab → reversed is ba. Not same. No.\n\n(0,2): a + aba → aaba → reversed is abaa. No.\n\n(1,0): b + a → ba → reversed ab. No.\n\n(1,2): b + aba → babaa → reversed aabab. No.\n\n(2,0): aba + a → abaa → reversed aaba. No.\n\n(2,1): aba + b → abab → reversed bababa? No. So none of these. So code returns No.\n\nSample input 3: two strings. Let's see:\n\nThe first string is 49 a's, the second is 50 a's. So when concatenated in either order?\n\nLet's see: for (0,1): 49 a's followed by 50 a's → total of 99 a's. Which is a palindrome. S",
    "### Question:\nYou are given a string S of length N consisting of lowercase English letters and the characters ( and ).\r\nPrint the string S after performing the following operation as many times as possible.\n\n- Choose and delete a contiguous substring of S that starts with (, ends with ), and does not contain ( or ) other than the first and last characters.\n\nIt can be proved that the string S after performing the operation as many times as possible is uniquely determined without depending on how it is performed.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nS\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 1 \\leq N \\leq 2 \\times 10^5\n- N is an integer.\n- S is a string of length N consisting of lowercase English letters and the characters ( and ).\n\nSample Input 1\n\n8\r\na(b(d))c\n\nSample Output 1\n\nac\r\n\nHere is one possible procedure, after which S will be ac.\n\n- Delete the substring (d) formed by the fourth to sixth characters of S, making it a(b)c.\n- Delete the substring (b) formed by the second to fourth characters of S, making it ac.\n- The operation can no longer be performed.\n\nSample Input 2\n\n5\r\na(b)(\n\nSample Output 2\n\na(\n\nSample Input 3\n\n2\r\n()\n\nSample Output 3\n\n\r\n\nThe string S after the procedure may be empty.\n\nSample Input 4\n\n6\r\n)))(((\n\nSample Output 4\n\n)))(((\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. The problem is to process a string by removing as many valid parentheses substrings as possible. The valid substrings here are those that start with '(', end with ')', and have no other parentheses in between. And we need to do this as many times as possible until no more operations can be done. The final string after all possible deletions is what we need to output.\n\nHmm. The key here is figuring out the order in which these substrings should be removed. But the problem says that the final result is unique regardless of the order. So maybe there's a way to find all the valid substrings that can be removed, and then remove them in a way that doesn't interfere with each other.\n\nWait, but when you remove a substring, that might affect other possible substrings. For example, if there's a nested structure like ( ( ) ), removing the inner first makes the outer valid, or vice versa. But according to the operation rules, the substring must start with '(' and end with ')', and have no other parentheses. So for the innermost parentheses, they are valid. Like in sample input 1: a(b(d))c. The (d) is innermost. Once removed, the (b ) becomes valid.\n\nSo perhaps the way to approach this is to find all the pairs of parentheses that form valid substrings and can be removed. But since removing some might allow others to become valid, how do we model this?\n\nAn approach that comes to mind is to process the string and find all the valid parentheses pairs that can be removed. But how?\n\nLet me think about using a stack. Stacks are commonly used for matching parentheses. For example, in standard matching problems, we push the index of '(', and when we find a ')', we pop and record a pair. However, in this problem, the requirement is that between the '(' and ')', there are no other parentheses. So each valid pair must have a matching '(' and ')', and their entire span contains no other parentheses. Which suggests that these pairs are \"atomic\" in some sense. \n\nWait, let's think: suppose we have a string like (a(b)c). The inner (b) is a valid substring. If we remove that first, then the outer becomes (a c), which has the letters a and c with no parentheses in between. So that would be a valid substring. So maybe the correct approach is to first remove the innermost valid pairs first. Because once they are removed, their positions are gone, allowing outer pairs to become valid.\n\nBut how to track which pairs are valid? Let me think again.\n\nThe stack approach for matching parentheses can be modified here. For example, when processing each character:\n\n- When we see '(', push its index onto the stack.\n- When we see ')', pop the last '(' index (if any) and form a pair. However, for this problem, the substring between these parentheses must not contain any other parentheses. Wait, no. Because when you pop, the current stack may indicate the last unmatched '('. But between that '(' and this ')', there could be other parentheses. For example, in ( () ), the inner pair is processed first. But when we process the outer ones, their indices would be recorded.\n\nWait, but how can we determine if a pair (i,j) (i is the index of '(', j is the index of ')') is such that between i and j, there are no other parentheses? Well, if we process the parentheses in the standard way with a stack, each time we find a ')', we can pair it with the most recent '('. And this pair would be such that between them, there are no other parentheses. Because if there were, they would have been processed first. Wait, no. For example, if we have a nested case: ( ( ) ). The first '(' is pushed, then the second '(', pushed. Then a ')', which pops the second '(', so that's a valid pair. Then another ')', which pops the first '(', so that's also a valid pair. But in this case, the outer pair includes the inner pair, which does have parentheses. So how can we find the valid substrings that can be removed?\n\nAh, the problem allows us to remove any valid substring that starts with '(' and ends with ')', with no other parentheses in between. So the innermost pairs are always valid. But when you have a pair that encloses other pairs, you can only remove it once all the enclosed pairs have been removed. So perhaps the way to model this is to find all the valid pairs (i.e., the ones that are properly matched) and then mark their positions. But then, when processing, which pairs are the ones that can be removed, considering that some are nested?\n\nAlternatively, perhaps we can model the process as follows: each time we find a valid pair (i.e., a pair of parentheses that directly enclose only letters, no other parentheses), we can remove them. But how to find those pairs?\n\nWait, but if we process the string once, removing such valid pairs, but after removal, new valid pairs may become available. For example, after removing inner pairs, their outer pairs may now be valid.\n\nThis suggests that we need to process the string iteratively. However, for large N (up to 2e5), an iterative approach that scans the string each time would be too slow.\n\nSo we need an efficient way to find all the pairs that can be removed in any order, considering that removing some allows others.\n\nLet me think again. How can we model the parentheses pairs that can be removed?\n\nLet me think of the problem as follows: the valid pairs are those that can be matched in such a way that between them, there are no other parentheses. For example, in the string \"a(b(d))c\", the (d) is such a pair. Once removed, the (b) becomes such a pair.\n\nSo perhaps we can model this with a stack. Let me think of the stack approach where we track the positions of '(', and when we find a ')', we check if there's a matching '('. But in addition, we can track for each pair whether between them there are other parentheses. Wait, but that's not directly possible.\n\nAlternatively, perhaps during the stack processing, each time we pair a '(' with a ')', we can record that pair. These pairs are properly matched. However, even if there are nested properly matched pairs inside, they would have been processed earlier. So for example, in the case of ( () ), the inner pair is recorded first. The outer pair is also recorded. Now, when considering which pairs to remove first, perhaps the inner ones are to be removed first. But how to track which pairs are allowed to be removed.\n\nBut the problem requires that during each operation, we delete a contiguous substring that starts with '(', ends with ')', and between them no other parentheses. So the inner pairs are exactly those that can be deleted immediately. Once they are deleted, ",
    "### Question:\nYou are given a length-N sequence A=(A_1,A_2,\\dots,A_N) consisting of 0, 1, and 2,\r\nand a length-N string S=S_1S_2\\dots S_N consisting of M, E, and X.\nFind the sum of\r\n\\text{mex}(A_i,A_j,A_k) over all tuples of integers (i,j,k) such that 1 \\leq i < j < k \\leq N and S_iS_jS_k= MEX.\r\nHere, \\text{mex}(A_i,A_j,A_k) denotes the minimum non-negative integer that equals neither A_i,A_j, nor A_k.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nA_1 A_2 \\dots A_N\r\nS\n\nOutput\n\nPrint the answer as an integer.\n\nConstraints\n\n\n- 3\\leq N \\leq 2\\times 10^5\n- N is an integer.\n- A_i \\in \\lbrace 0,1,2\\rbrace\n- S is a string of length N consisting of M, E, and X.\n\nSample Input 1\n\n4\r\n1 1 0 2\r\nMEEX\n\nSample Output 1\n\n3\r\n\nThe tuples (i,j,k)\\ (1 \\leq i < j < k \\leq N) such that S_iS_jS_k = MEX are the following two: (i,j,k)=(1,2,4),(1,3,4).\r\nSince \\text{mex}(A_1,A_2,A_4)=\\text{mex}(1,1,2)=0 and \\text{mex}(A_1,A_3,A_4)=\\text{mex}(1,0,2)=3, the answer is 0+3=3.\n\nSample Input 2\n\n3\r\n0 0 0\r\nXXX\n\nSample Output 2\n\n0\n\nSample Input 3\n\n15\r\n1 1 2 0 0 2 0 2 0 0 0 0 0 2 2\r\nEXMMXXXEMEXEXMM\n\nSample Output 3\n\n13\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this programming problem. Let me read the problem carefully.\n\nSo the task is: Given a sequence A of 0,1,2 and a string S of M, E, X, find the sum of mex of all triplets (i,j,k) where i<j<k, the corresponding S characters are M, E, X in order, and then compute mex of their A values.\n\nThe constraints are N up to 2e5. So O(N^3) is impossible. Need an efficient approach.\n\nFirst, the mex function. Since each A_i is 0,1,2, the mex can be 0,1,2,3. Let me think how to compute it. For example:\n\nIf the triplet has values (1,1,2), then the mex is 0, since 0 is missing. For (1,0,2), all 0,1,2 are present, mex is 3.\n\nSo for any triplet, the mex depends on which of 0,1,2 are present. Let me think:\n\nThe mex is the smallest non-negative integer not in the triplet's values.\n\nSo for a triplet (a,b,c):\n\nCheck if 0 is missing → mex is 0.\n\nElse check if 1 is missing → mex is 1.\n\nElse check if 2 is missing → mex is 2.\n\nElse mex is 3.\n\nBut since each A_i can be 0,1,2, how can we compute this quickly?\n\nNow, the main challenge is to find all (M, E, X) triplets in S, where i<j<k, and for each such triplet, compute mex of A_i, A_j, A_k, then sum all those mex values.\n\nThe problem is how to efficiently count all such MEX patterns and compute their mex contributions.\n\nThe naive approach would be to iterate over all possible i<j<k, check if S has MEX, and if so, compute mex. But for N=2e5, there are O(N^3) possible triplets, which is way too big. So we need a smarter way.\n\nBut since S needs to be M followed by E followed by X, with i<j<k, perhaps we can precompute for each E in the string, how many Ms are before it, and how many Xs are after it. Then for each E, the number of triplets that use this E is (number of Ms before) * (number of Xs after). But that's for just counting the number of valid triplets. But here, for each such triplet (M, E, X), we need to consider their A values and calculate mex. However, the A values can vary for each combination. So how can we track the counts of different combinations?\n\nSo, perhaps for each E in the string, we can look at all possible Ms before it and Xs after it. Then for each such triplet, we need to collect the A values of M, E, X, compute mex, and sum all of them.\n\nBut again, if we do this naively, for each E, if there are M Ms and X Xs, it could be O(M*X) per E, leading to O(N^2) time in total. Which is not feasible for N=2e5. So we need a way to calculate for each E, the total contribution from all Ms before and Xs after, without iterating through all possible pairs.\n\nLet me think: For each occurrence of 'E' in S, let's denote:\n\n- The count of 'M's before this E: m_count.\n\nBut also, for those Ms before, we have their A values (either 0, 1, or 2). Similarly, for Xs after this E, their A values are also 0, 1, or 2.\n\nSo for each E, if we can know, for the Ms before, how many have A=0, A=1, A=2. Similarly for Xs after, how many have A=0, A=1, A=2. Then, for each E, we can compute the total contribution by considering all combinations of M's A, E's A, and X's A.\n\nBecause for each triplet (M, E, X), the mex is determined by the triplet (a_m, a_e, a_x). So for each E, if we know how many Ms before have each possible a_m, and how many Xs after have each possible a_x, then for each possible combination of a_m and a_x, we can calculate how many times each mex occurs, multiplied by the count of Xs and Ms.\n\nSo here's the plan:\n\nLoop through each position j where S[j] is 'E'. For each such j:\n\n- Let a_e be A[j].\n\n- Compute the count of Ms before j, grouped by their A value. Let's say, m0 is the number of Ms before with A_i=0, m1 for A_i=1, m2 for A_i=2.\n\n- Compute the count of Xs after j, grouped by their A value. Let's say x0, x1, x2.\n\nThen, for all possible combinations of (a_m, a_x), we can compute how many triplets (M, E, X) contribute to the sum. For each a_m (0,1,2) and a_x (0,1,2), the contribution would be (count of a_m) * (count of a_x) multiplied by the mex(a_m, a_e, a_x).\n\nSo for each E, we need to:\n\n- For all a_m in 0,1,2 and a_x in 0,1,2:\n\n   if there are c = (m_a_m) * (x_a_x) such triplets, then add c * mex(a_m, a_e, a_x) to the total sum.\n\nSo, how to compute m0, m1, m2 for each E? We can pre-process for each position, the cumulative counts of M's with A=0, etc. For example, create a prefix array that for each index i, stores the number of M's up to i-1, along with their A values counts. Similarly for X's, but it's suffix arrays.\n\nBut how?\n\nLet me think about the prefix part. For each position j, if we want to know, in the positions before j (i < j), how many Ms there are with A=0, A=1, A=2. So we can create a prefix array where prefix_m[i][v] represents the count of Ms in S[0..i) (up to but not including i) with A value v.\n\nSimilarly for the suffix part: for each position j, how many Xs are there after j (k > j) with each A value. So suffix_x[i][v] represents the count of Xs in S[i..n-1] (from i onwards) ?\n\nSo for example, for the prefix part, we can have for each index i (from 0 to N), the counts of M's up to i-1. Similarly for the suffix.\n\nLet me think of the steps:\n\nFirst, precompute prefix_m_counts, which is an array of size N+1 (so that for index j, we can have up to j Ms considered). Each entry is a dictionary or three numbers: count of 0s, 1s, 2s among the Ms in the prefix up to index j.\n\nHow to build it? For each position from 0 to N-1:\n\nprefix_m_counts[j+1] = prefix_m_counts[j].copy()\n\nBut if S[j] is 'M', then for the current A[j], we add 1 to the corresponding count.\n\nSame for others.\n\nSimilarly for suffix_x_counts. For each position j, the suffix_x_counts[j] would represent the counts of Xs from j to the end.\n\nBut how to compute that? We can do a reverse pass. For each j from N-1 down to 0:\n\nsuffix_x_counts[j] = suffix_x_counts[j+1].copy()\n\nBut if S[j] is 'X', then add 1 to the corresponding A[j] count.\n\nOnce these prefix and suffix arrays are built, then for each position j where S[j] is 'E', we can get:\n\nm0 = prefix_m_counts[j]['M' counts with A=0? Wait, no. The prefix_m_counts[j] holds the counts of Ms before j (since for the j-th position in the prefix array, it's up to j elements, but for the j-th position in S, the prefix up to j is up to index j in the prefix array. Let me think:\n\nLet me index S as 0-based. For j in 0..N-1:\n\nprefix_m_counts[j+1][a] = prefix_m_counts[j][a] + (if S[j] is 'M', then 1 if A[j] is a, else 0).\n\nSo, for a given j (the position of 'E'), the number of Ms before j (i.e., positions 0 to j-1) is in prefix_m_counts[j].\n\nSimilarly, for the Xs after j (positions j+1 to N-1). For that, suffix_x_counts[j] would hold the counts of Xs in positions >=j? Or maybe we need to compute for each j, the number of Xs in positions after j (so j+1 to end). Let me think:\n\nWe can precompute a suffix array wh",
    "### Question:\nThere are N people numbered 1 to N.\r\nYou are given their schedule for the following D days. The schedule for person i is represented by a string S_i of length D. If the j-th character of S_i is o, person i is free on the j-th day; if it is x, they are occupied that day.\nFrom these D days, consider choosing some consecutive days when all the people are free.\r\nHow many days can be chosen at most? If no day can be chosen, report 0.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN D\r\nS_1\r\nS_2\r\n\\vdots\r\nS_N\n\nOutput\n\nPrint the maximum number of days that can be chosen, or 0 if no day can be chosen.\n\nConstraints\n\n\n- 1 \\leq N \\leq 100\n- 1 \\leq D \\leq 100\n- N and D are integers.\n- S_i is a string of length D consisting of o and x.\n\nSample Input 1\n\n3 5\r\nxooox\r\noooxx\r\noooxo\n\nSample Output 1\n\n2\r\n\nAll the people are free on the second and third days, so we can choose them.\r\nChoosing these two days will maximize the number of days among all possible choices.\n\nSample Input 2\n\n3 3\r\noxo\r\noxo\r\noxo\n\nSample Output 2\n\n1\r\n\nNote that the chosen days must be consecutive. (All the people are free on the first and third days, so we can choose either of them, but not both.)\n\nSample Input 3\n\n3 3\r\noox\r\noxo\r\nxoo\n\nSample Output 3\n\n0\r\n\nPrint 0 if no day can be chosen.\n\nSample Input 4\n\n1 7\r\nooooooo\n\nSample Output 4\n\n7\n\nSample Input 5\n\n5 15\r\noxooooooooooooo\r\noxooxooooooooox\r\noxoooooooooooox\r\noxxxooooooxooox\r\noxooooooooxooox\n\nSample Output 5\n\n5\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's see. I need to solve this programming problem. Let me read the problem carefully again.\n\nSo, the problem is about finding the maximum number of consecutive days where all N people are free. Each person's schedule is given as a string of 'o' and 'x' for each day. The task is to determine the longest such consecutive sequence.\n\nHmm. The approach here is to check for each possible consecutive sequence of days, and find the maximum length where all people have 'o' on all those days.\n\nFirst, let's think about how to represent the data. For each day, we need to know if all people are free. Let's think of each day as a column in the strings. For each column j (from 0 to D-1, maybe), check if all the S_i have 'o' in that column. Then, the problem becomes finding the maximum number of consecutive 1s in a binary array where 1 indicates all people are free that day, and 0 indicates at least one person is busy.\n\nYes. So the steps could be:\n\n1. For each day (from 0 to D-1), determine if all people are free. Let's create an array called all_free where all_free[j] is True if all people are free on day j+1, else False.\n\nWait, but how do we check that for each day? For each day j (0-based?), check for every person i from 0 to N-1, whether S_i[j] is 'o'. If all are 'o', then that day is a candidate.\n\nOnce we have this array, the problem reduces to finding the maximum length of consecutive True values in this array.\n\nYes. Then the problem becomes similar to the classic \"longest run of 1s\" problem. For example, if all_free is [True, True, False, True, True, True, False], then the maximum is 3.\n\nSo the plan is:\n\n- Read N and D.\n- Read N strings, each of length D.\n- For each day (column) in 0 to D-1:\n   - Check if all N people have 'o' in that column.\n   - Record as 1 (if yes) or 0 (otherwise) in an array.\n- Now find the maximum number of consecutive 1s in this array.\n\nWait, but how do I process the days? For example, for the sample input 1:\n\nSample Input 1:\n\n3 5\n\nxooox\n\noooxx\n\noooxo\n\nThe days are 5 days. For each day:\n\nDay 1: check all people's first character.\n\nPerson 1: x (not free), so day 1 is not all free. Similarly:\n\nLet me think. The first string is xooox. So days 0 is 'x', days 1 is 'o', 2 'o', 3 'o', 4 'x'.\n\nSecond string is oooxx. So day 0 is 'o', day 4 is 'x'.\n\nThird string is oooxo. So day 0 is 'o', day 4 is 'o'.\n\nSo for each day:\n\nday 0: x (person 0 is busy. So all_free[0] is false.\n\nday 1: all have 'o'? Let's check each person's day 1. Person 0: o, person 1: o, person 2: o. So yes. So all_free[1] is True.\n\nday 2: same. All o?\n\nPerson 0: o, person 1: o, person 2: o. Yes.\n\nday 3: person 0: o, person 1: x (since the fourth character of the second string is 'x'? Wait let me check.\n\nWait the second string is \"oooxx\". For day 3 (assuming 0-based), the index is 3. Let's see the string:\n\nIndices 0: o, 1: o, 2: o, 3: x, 4: x. Oh right. So for day 3, person 1 has x. So all_free[3] would be false.\n\nSo the all_free array for sample input 1 would be:\n\n[False, True, True, False, False]. Then the maximum consecutive is 2. Which matches sample output 1.\n\nYes. So then the problem is to compute this array and find the maximum consecutive 1s.\n\nSo the steps are:\n\nHow to generate the all_free array:\n\nFor each day j in 0..D-1:\n\nCheck for all i in 0..N-1, S_i[j] is 'o'. If all are, then mark as 1, else 0.\n\nOnce that array is generated, compute the maximum consecutive 1s.\n\nNow, how to compute the maximum consecutive 1s?\n\nWe can iterate through the array, keeping a current counter and a maximum counter.\n\nInitialize current_streak and max_streak to 0.\n\nFor each element in the array:\n\n   if element is 1 (True), increment current_streak by 1.\n\n   else, reset current_streak to 0.\n\n   update max_streak as the maximum between max_streak and current_streak.\n\nSo for example, for the array [0,1,1,0,0], the max_streak would be 2.\n\nYes.\n\nNow, coding steps:\n\nRead N and D.\n\nThen read N strings. Each string is of length D.\n\nThen, for each day j from 0 to D-1:\n\n   check for all S_i[j] == 'o' for all i.\n\nHow to check that? For each j, loop over all i from 0 to N-1, check if any is 'x'. If any is 'x', then day j is not all free.\n\nAlternatively, for each day j, check all the characters in position j of all the strings. If all are 'o', then it's a good day.\n\nSo in code:\n\nall_free = []\n\nfor j in 0 to D-1:\n\n   is_all_free = True\n\n   for i in 0 to N-1:\n\n      if S_i[j] != 'o':\n\n          is_all_free = False\n\n          break\n\n   all_free.append(1 if is_all_free else 0)\n\nOnce that's done, compute the maximum consecutive 1s.\n\nThen output that number.\n\nYes.\n\nNow, code in Python:\n\nRead N and D from input.\n\nThen read N lines, each line being S_i.\n\nThen process each day.\n\nSo for example, for D days.\n\nNow, code:\n\nRead N, D = map(int, input().split())\n\nThen schedules = [input().strip() for _ in range(N)]\n\nThen create the all_free array as a list of 0 and 1.\n\nBut perhaps better to create a list of booleans, then convert to 0/1 for the streak calculation.\n\nWait, actually, for the streak calculation, we can just check if the day is all free, and increment accordingly.\n\nSo code:\n\nmax_streak = 0\n\ncurrent = 0\n\nfor j in 0 to D-1:\n\n   if all_free_day(j):\n\n       current += 1\n\n       if current > max_streak:\n\n           max_streak = current\n\n   else:\n\n       current = 0\n\nBut how to implement all_free_day?\n\nSo, for day j (0-based?), check for each string in schedules, the j-th character.\n\nSo:\n\ndef is_all_free(j):\n\n    for s in schedules:\n\n        if s[j] != 'o':\n\n            return False\n\n    return True\n\nBut in Python, strings are zero-based, and the days are from 0 to D-1.\n\nYes.\n\nSo putting it all together.\n\nWait, but what if D is 0? Well, constraints say D is at least 1? No, looking at sample input 4, D can be 7, but the constraints say D >=1?\n\nWait the problem says:\n\nConstraints:\n\n1 ≤ N ≤ 100\n\n1 ≤ D ≤ 100\n\nYes, so D is at least 1. So no problem.\n\nSo code:\n\nRead N, D.\n\nschedules = [input().strip() for _ in range(N)]\n\nThen:\n\ncurrent_streak = 0\n\nmax_streak = 0\n\nfor day in 0 to D-1:\n\n   all_free = True\n\n   for person in 0 to N-1:\n\n       if schedules[person][day] != 'o':\n\n           all_free = False\n\n           break\n\n   if all_free:\n\n       current_streak += 1\n\n       if current_streak > max_streak:\n\n           max_streak = current_streak\n\n   else:\n\n       current_streak = 0\n\nAt the end, output max_streak.\n\nWait, but in code, how to loop through days? For day in range(D):\n\nYes.\n\nSo that's ",
    "### Question:\nThere is a grid with H rows and W columns. Let (i, j) denote the square at the i-th row from the top and j-th column from the left of the grid.\r\nEach square of the grid is holed or not. There are exactly N holed squares: (a_1, b_1), (a_2, b_2), \\dots, (a_N, b_N).\nWhen the triple of positive integers (i, j, n) satisfies the following condition, the square region whose top-left corner is (i, j) and whose bottom-right corner is (i + n - 1, j + n - 1) is called a holeless square.\n\n- i + n - 1 \\leq H.\n- j + n - 1 \\leq W.\n- For every pair of non-negative integers (k, l) such that 0 \\leq k \\leq n - 1, 0 \\leq l \\leq n - 1, square (i + k, j + l) is not holed.\n\nHow many holeless squares are in the grid?\n\nInput\n\nThe input is given from Standard Input in the following format:\nH W N\r\na_1 b_1\r\na_2 b_2\r\n\\vdots\r\na_N b_N\n\nOutput\n\nPrint the number of holeless squares.\n\nConstraints\n\n\n- 1 \\leq H, W \\leq 3000\n- 0 \\leq N \\leq \\min(H \\times W, 10^5)\n- 1 \\leq a_i \\leq H\n- 1 \\leq b_i \\leq W\n- All (a_i, b_i) are pairwise different.\n- All input values are integers.\n\nSample Input 1\n\n2 3 1\r\n2 3\n\nSample Output 1\n\n6\r\n\nThere are six holeless squares, listed below. For the first five, n = 1, and the top-left and bottom-right corners are the same square.\n\n- The square region whose top-left and bottom-right corners are (1, 1).\n- The square region whose top-left and bottom-right corners are (1, 2).\n- The square region whose top-left and bottom-right corners are (1, 3).\n- The square region whose top-left and bottom-right corners are (2, 1).\n- The square region whose top-left and bottom-right corners are (2, 2).\n- The square region whose top-left corner is (1, 1) and whose bottom-right corner is (2, 2).\n\nSample Input 2\n\n3 2 6\r\n1 1\r\n1 2\r\n2 1\r\n2 2\r\n3 1\r\n3 2\n\nSample Output 2\n\n0\r\n\nThere may be no holeless square.\n\nSample Input 3\n\n1 1 0\n\nSample Output 3\n\n1\r\n\nThe whole grid may be a holeless square.\n\nSample Input 4\n\n3000 3000 0\n\nSample Output 4\n\n9004500500\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to count the number of holeless squares in a grid. Let's think about how to approach this.\n\nFirst, the problem says that a holeless square is a square region that doesn't contain any holed squares. The grid has H rows and W columns. There are N holed squares given as input.\n\nThe straightforward way would be to check every possible square of every possible size. But with H and W up to 3000, that's going to be way too slow. Because for each possible top-left corner, checking all possible n (size) up to the maximum possible, and for each n checking all the cells in the square. That would be O(H*W*min(H,W)^2) time, which is definitely not feasible for 3000.\n\nSo I need a more efficient approach. Let me think about dynamic programming or some pre-processing.\n\nWait, there's a classic problem of counting the number of squares in a grid. For that problem, if you have a grid without any holes, the total number is sum(i*(W - i) for i in 1..min(H,W)), but here we have holes to consider.\n\nBut how do the holes affect the count here?\n\nAnother idea: For each cell (i,j), determine the largest possible square size that can have (i,j) as the top-left corner. Then, the number of squares contributed by (i,j) is equal to that maximum size. But how to compute this efficiently when there are holes?\n\nIn the standard problem without holes, we use dynamic programming where dp[i][j] represents the size of the largest square with (i,j) as the bottom-right corner. But here, perhaps we can adjust that approach.\n\nAlternatively, maybe we can pre-process the grid to know for each cell the nearest hole to the right and below, but I'm not sure.\n\nWait, but the problem requires that the entire square has no holes. So for a square starting at (i,j) of size n, all cells in that n x n square must be non-holed.\n\nSo perhaps we can model for each cell (i,j) the maximum possible n such that starting from (i,j), there's a square of size n with no holes. Then the total number of squares is the sum of these n's for all (i,j).\n\nThe challenge is to compute this maximum n for each cell efficiently.\n\nHow to compute this maximum n for each cell?\n\nAn approach inspired by the classic DP method could be used here. Let's think: For a cell (i,j), if it's a holed cell, then the maximum n is 0. Otherwise, the maximum n is 1 plus the minimum of the maximum ns of (i+1,j), (i,j+1), and (i+1,j+1). But wait, that's the standard approach for finding the largest square of 1s in a binary matrix. However, in our case, the problem is inverted: holes are like 1s (bad) and non-holes are 0s (good)? Or maybe the other way around.\n\nWait, let me think again. Let me imagine that the grid is such that a cell is 1 if it's not holed. Then, the problem becomes finding all possible squares of 1s, but here we need to count all possible squares. However, the standard DP approach for largest square can be adapted here.\n\nBut the standard DP approach for the largest square gives for each cell (i,j) the size of the largest square with (i,j) as the bottom-right corner. However, in our problem, perhaps we need to compute for each cell (i,j) the maximum possible square size that has (i,j) as the top-left corner. Which is different.\n\nHmm. How to compute that?\n\nAlternatively, perhaps for each cell (i,j), we can find the maximum n such that there's a square of size n starting here. How to compute that?\n\nLet me think for each cell (i,j). Suppose we process the grid in some order. For example, for each cell (i,j), we can track the maximum consecutive non-holed cells to the right and down. But how does that help for squares?\n\nAnother idea: For each cell (i,j), if it's not holed, then the maximum square size starting here is 1 plus the minimum of the maximum square sizes starting at (i+1,j), (i,j+1), and (i+1,j+1). Wait, that sounds familiar. Let me think. Suppose we have a square of size k starting at (i,j). Then, the square can be divided into three adjacent squares: the cell to the right (which would be a square of size k-1 starting at (i,j+1)), the cell below (square of size k-1 starting at (i+1,j)), and the square to the bottom-right (square of size k-1 starting at (i+1,j+1)). Therefore, if all those three positions can support a square of size k-1, then (i,j) can support a square of size k. Therefore, the recurrence would be:\n\ndp[i][j] = 1 + min(dp[i+1][j], dp[i][j+1], dp[i+1][j+1])\n\nBut this recurrence requires that (i,j) is not holed, and that the three neighboring cells are also part of valid squares.\n\nBut this recurrence is processed in reverse order. For example, we need to process the grid from bottom to top and right to left.\n\nThis seems promising. Let's test this with sample input 1.\n\nSample Input 1:\n\nH=2, W=3, N=1. The holed cell is (2,3).\n\nThe grid is:\n\nRow 1: columns 1,2,3 (all non-holed except none?), except (2,3). So the grid for holed cells is:\n\nHoles are at (2,3).\n\nSo the dp array would be processed starting from bottom-right.\n\nLet me imagine the grid:\n\nFor cells (i,j), dp[i][j] represents the maximum square size that can start at (i,j).\n\nLet me compute for each cell.\n\nLet me consider H=2 rows, W=3 columns.\n\nLet me index rows from 1 to H, columns 1 to W.\n\nWe process from i=H downto 1, and j=W downto 1.\n\nStarting with cell (2,3): since it's holed, dp[2][3] = 0.\n\nCell (2,2): not holed. So check cells (2+1, 2) which is out of bounds (i+1=3>2). Similarly for (2,2+1=3) and (2+1, 2+1=3). Since those are beyond the grid, their dp values can be considered as 0. So min of (dp[3][2] (0), dp[2][3] (0), dp[3][3] (0)) → min is 0. So dp[2][2] is 1 + 0 = 1. So there's a square of size 1 here.\n\nSimilarly for cell (2,1): check (3,1), (2+1, j+1?), etc. Since those are out of bounds, their dp is 0. So dp[2][1] is 1.\n\nFor cell (1,3): check (2,3) (dp is 0), (1+1, j=3), (i, j+1) is (1,4) which is invalid. So min( dp[2][3]=0, ... others 0 → min is 0. So dp[1][3] is 1.\n\nNow cell (1,2): check (i+1,j) → (",
    "### Question:\nYou are given a non-empty string S consisting of (, ), and ?.\r\nThere are 2^x ways to obtain a new string by replacing each ? in S with ( and ), where x is the number of occurrences of ? in S.  Among them, find the number, modulo 998244353, of ways that yield a parenthesis string.\nA string is said to be a parenthesis string if one of the following conditions is satisfied.\n\n- It is an empty string.\n- It is a concatenation of (, A, and ), for some parenthesis string A.\n- It is a concatenation of A and B, for some non-empty parenthesis strings A and B.\n\nInput\n\nThe input is given from Standard Input in the following format:\nS\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- S is a non-empty string of length at most 3000 consisting of (, ), and ?.\n\nSample Input 1\n\n(???(?\n\nSample Output 1\n\n2\r\n\nReplacing S with ()()() or (())() yields a parenthesis string.\r\nThe other replacements do not yield a parenthesis string, so 2 should be printed.\n\nSample Input 2\n\n)))))\n\nSample Output 2\n\n0\n\nSample Input 3\n\n??????????????(????????(??????)?????????(?(??)\n\nSample Output 3\n\n603032273\r\n\nPrint the count modulo 998244353.\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to count the number of valid parenthesis strings that can be formed by replacing the '?' in the given string. The answer should be modulo 998244353. Let's think about how to approach this.\n\nFirst, I remember that valid parentheses require that at every point in the string, the number of opening brackets is not less than the closing ones, and overall they are equal. But since there are question marks, which can be either, this seems like a dynamic programming problem.\n\nDynamic Programming approach: Let's consider dp[i][j], which represents the number of valid ways to form a string up to some position i, with a balance of j open brackets. Balance here means the number of '(' minus the number of ')'. For a valid string, this balance should never drop below zero, and end at zero.\n\nThe initial state would be dp[0][0] = 1, since before processing any characters, the balance is zero.\n\nFor each position in the string (let's say we're at the ith character, 0-based or 1-based?), and for each possible balance j, we transition based on the current character.\n\nLet me think about the transitions. Suppose we are at position i, considering the ith character. For each possible balance j before processing this character, we want to see what choices we can make for this character and what the new balance would be.\n\nIf the current character is '(', then the balance increases by 1. If it's ')', the balance decreases by 1 (so j must be at least 1 to allow this). If it's '?', then we consider both possibilities.\n\nBut since the string must be valid at every step, we need to ensure that during processing, the balance never becomes negative.\n\nLet me formalize this. For the DP state:\n\ndp[i][j] is the number of ways to process the first i characters such that the balance is j. Then, for each i and j, we can compute dp[i+1][new_j] based on the current character.\n\nFor example, for the current character S[i], when processing from dp[i][j], we can:\n\nCase 1: S[i] is '(':\n\nnew_balance = j + 1. So, dp[i+1][j+1] += dp[i][j]\n\nCase 2: S[i] is ')': \n\nnew_balance = j - 1. However, for this to be allowed, j must be at least 1 (since balance can't be negative). \n\nCase 3: S[i] is '?', then both possibilities are allowed. So we add both cases.\n\nNow, the constraints for the DP transitions are:\n\nAt any step, after processing, the balance must be non-negative. Also, the maximum possible balance depends on the number of characters processed so far. But for the DP, we can track all possible balances up to some maximum value.\n\nNow, the question is, what's the maximum possible balance? For a string of length n, the maximum possible balance is n (if all characters are '('). But since we need to make sure that eventually, the balance ends at zero, during processing the balance can't exceed n - i, where i is the current position. But for DP purposes, perhaps we can track up to the maximum possible balance at each step.\n\nThe length of the string is up to 3000, so for each i (up to 3000), j can be up to some value. For example, after processing i characters, the maximum possible balance is i (if all were '('). But since we need the total number of parentheses to be balanced, at the end (after processing all characters), the balance must be zero. Therefore, during processing, for the balance to have a chance to return to zero, at any step i, the balance j can't exceed (n - i), where n is the total length. Otherwise, even if all remaining characters are ')', it's impossible to bring the balance back down. So during the DP transitions, j can range from 0 up to some maximum value, which for each step i is up to i (since each character contributes at most +1 or -1).\n\nSo, for the DP table, the dimensions would be (n+1) x (max_balance + 1). Since n can be up to 3000, the max_balance is 3000. So the DP table is manageable for 3000x3000, which is 9 million entries. That's feasible in Python with proper optimization.\n\nNow, let's outline the steps:\n\nInitialize a 2D array dp where dp[i][j] represents the number of valid ways after processing i characters, with a balance of j.\n\nInitialize dp[0][0] = 1.\n\nFor each i from 0 to n-1 (n is the length of S):\n\n   for each possible balance j at this step:\n\n      if S[i] is '(':\n\n          new_j = j + 1\n\n          if new_j <= current possible max (which is i+1, but we need to check that it's not too big for future steps. However, since we track up to possible balances, and for each step, the maximum j is up to i+1, but in the DP, we can just allow it as long as it's non-negative and not exceeding some maximum possible value that allows it to be balanced eventually. However, during processing, even if it exceeds, but there's no way to balance it, then those paths will not contribute to the final answer. So during transitions, we can allow j up to some maximum value. So for each i, the maximum j can be up to i+1, but during the DP steps, we can just process all possible j's up to n (since n is 3000).\n\n      So, for each possible j, if the current character is '(', then add dp[i][j] to dp[i+1][j+1].\n\n      Similarly for ')', check if the current j can subtract 1, and add to dp[i+1][j-1].\n\n      For '?', do both possibilities.\n\nBut also, during processing, the balance must never be negative. So during transitions, when processing a character, we have to ensure that the new balance is not negative. For example, adding a ')' would decrease the balance by 1, so we can only do that if the current balance is at least 1.\n\nSo, for each transition:\n\nIf current character is '(':\n\nnew_j = j + 1. Since adding a '(', so balance increases. No risk of negative here. So we can add to dp[i+1][new_j].\n\nIf current character is ')':\n\nWe can only proceed if j >= 1. Then new_j is j-1. \n\nIf current character is '?', then:\n\nwe have two possibilities: either it's '(', contributing to new_j = j+1, or ')', contributing to new_j = j-1 (but only if j >=1).\n\nSo during the DP transitions, for each case, we have to check these conditions.\n\nOnce all characters are processed, the answer is dp[n][0], since after processing all n characters, the balance must be 0.\n\nBut wait, also during the processing, the balance must never drop below zero. So for example, if during transitions, adding a ')' leads to a negative balance, that path is invalid and should not be counted. Hence, in the DP transitions, when processing ')', we can only proceed if j >= 1, because otherwise j-1 would be negative.\n\nTherefore, the DP transitions automatically enforce that the balance never drops below zero. Because for each step, when considering transitions, we only allow moves that do not lead to negative balance.\n\nIs that sufficient? Let me think. Suppose during processing, the balance is j. For the next step, whe",
    "### Question:\nThe number pi to the 100-th decimal place is\n3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679.\nYou are given an integer N between 1 and 100, inclusive.\nPrint the value of pi to the N-th decimal place.\nMore precisely, truncate the value of pi to N decimal places and print the result without removing the trailing 0s.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\n\nOutput\n\nPrint the value of pi to the N-th decimal place in a single line.\n\nConstraints\n\n\n- 1\\leq N\\leq 100\n- N is an integer.\n\nSample Input 1\n\n2\n\nSample Output 1\n\n3.14\n\nTruncating the value of pi to 2 decimal places results in 3.14. Thus, you should print 3.14.\n\nSample Input 2\n\n32\n\nSample Output 2\n\n3.14159265358979323846264338327950\n\nDo not remove the trailing 0s.\n\nSample Input 3\n\n100\n\nSample Output 3\n\n3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, I need to solve this problem where I have to print the value of pi truncated to N decimal places, including any trailing zeros. Let's see.\n\nThe problem says that pi is given up to the 100th decimal place. So the input is an integer N between 1 and 100. My task is to output pi truncated to exactly N decimals. Truncate means I don't round it, just cut off after the Nth digit. And I must not remove the trailing zeros. \n\nFirst, I need to store the full value of pi up to 100 decimals as a string. Let me check the sample inputs. For example, when N is 2, the output is 3.14. For N=32, it's the first 32 decimal digits. So the key is to take the first N digits after the decimal point, without rounding.\n\nSo the first step is to have a string that represents pi with exactly 100 decimals. The given value is:\n\n3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679.\n\nWait, let me count. Let me check how many digits are there after the decimal. Let's see:\n\nThe string starts with \"3.\" followed by 100 digits. Let me confirm. For example, sample input 3 is N=100, and the output is that entire string. So the given pi value has exactly 100 decimal digits.\n\nSo my plan is:\n\n1. Store the entire pi string as a constant in the code. Let me copy that string exactly. Wait, the given pi string ends with ...679. So the full string is 3. followed by 100 digits. So the code can have a variable like pi_str that holds this string.\n\nThen, given N, I need to take the substring up to N digits after the decimal point, and output that. But how?\n\nFor example, the structure of the pi_str is \"3.1415926535...\" where after the decimal point there are 100 digits. So for N=2, we take the first two digits after the decimal. So the output is \"3.14\".\n\nSo the approach is:\n\n- Take the first (N+1) characters? Wait, because the integer part is 3., then the decimal part. So the total length of the string is 1 (the 3) + 1 (the decimal point) + 100 digits. So for N decimal places, the total length of the string should be 1 + 1 + N digits. Wait, no. Because for example, for N=2, the output is \"3.14\" which is 1 (3) + 1 (.) + 2 digits. So total length is 4 characters.\n\nSo the pi string given in the problem is already correctly structured. So the code can take the first (2 + N) characters? Let's see.\n\nWait, the original pi string is 3. followed by 100 digits. So the index of the decimal point is 1. Let's think about how to slice this. Suppose the string is s. Then, for N decimal digits, the required string is the first (N+2) characters? Let me check.\n\nFor example, if N is 2: the string starts with \"3.14...\", so taking first 2 + 1 (the 3 and the decimal point) plus 2 digits? Wait, 3.14 is 4 characters. Which is 2 digits after decimal. So the length is 1 (the 3) + 1 (the .) + N digits. So total length is 2 + N. So for N=2, the length is 4. So the original pi string is 3. followed by 100 digits. So the first (2 + N) characters would be correct. Let me check.\n\nOriginal string: \"3.1415926535...\" for N=2, take first 2+2=4 characters? Let's see: 3.14. Yes. So for general N, the answer is the first (1 + 1 + N) = N+2 characters of the pi_str.\n\nBut wait, what if N is 100? The entire string is 3. followed by 100 digits. So the length of the string is 1 + 1 + 100 = 102 characters. So for N=100, the required output is 3. followed by 100 digits. So taking the first 100 + 2 = 102 characters. Which is the entire string. Which is correct.\n\nYes! So the solution is to take the first (N + 2) characters of the given pi string. Because the pi string in the problem includes exactly 100 decimal digits. So the code can do something like:\n\npi_str = \"3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679\"\n\nThen, for input N, output the first N+2 characters.\n\nBut wait, how about when N is 0? Well, the problem says N is between 1 and 100 inclusive. So no problem.\n\nSo the steps are:\n\n1. Read N from input.\n\n2. Take the first N+2 characters of the pi_str.\n\nBut how to get the pi_str correctly?\n\nWait, the sample input for N=100 gives the output as the entire string. Let's check:\n\nOriginal pi_str (as given) is:\n\n3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679\n\nWhich is 1 (3) + 1 (.) + 100 digits. Let's count the digits in the sample output 3. Let me check the sample input 3's output. The output is 3. followed by 100 digits. So the length of the output is 102 characters. So for N=100, our code would take N+2=102 characters. Which is exactly that.\n\nSo the code's pi_str must be exactly the given string, but how is it written?\n\nWait, in the problem statement, the given pi to 100 decimals is written as:\n\n3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679.\n\nNote the trailing period? Or is that a typo? Wait, looking at the problem statement's sample input 3's output. The sample input 3 is N=100, and the output is that long number. Let me check the sample input 3's output. The given output ends with ...679. Let me check the pi_str given in the problem. Oh, the problem says that the value of pi to 100 decimals is that string. So the string is 3. followed by 100 digits, but in the problem's description, perhaps the trailing '.' is a typo or not? Wait, no. Let me check the problem's sample input 3's output. The sample output for N=100 is that long string ending with 679. Let me count the digits in the sample output 3.\n\nBut perhaps the problem's given pi string in the first line is missing a '.' at the end? Or maybe it's a typo. For example, in the problem statement, the given pi string is written as ending with ...679. But if that's the case, then the actual correct pi digits up to 100 would be as given. So perhaps in the code, the correct pi string is stored as the given one. So I need to make sure that the code's pi_str is correct.\n\nBut how? Well, the sample outputs are the best source. For example, sample input 1 is N=2, output is 3.14. Let's check the code. Let me assume that the given pi_str is correct.\n\nSo the code can be structured as follows:\n\nRead N.\n\nTake the first N+2 characters of the pi_str and print them.\n\nNow, the code must have the correct pi_str. So I need to copy the exact string from the problem statement. But in the problem statement, the given pi string is written as:\n\n3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679.\n\nWait, perhaps the problem statement's first line ends with a '.'? Or that's a typo. Let me think. For sample input 3, when N=100, the output is that string without the trailing '.'? Because in the problem statement, the sample output 3 is:\n\n3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679\n\nWhich is exactly the same as the given pi string in the problem. Except that perhaps the pro",
    "### Question:\nN players, player 1, player 2, ..., player N, participate in a game tournament. Just before the tournament starts, each player forms a one-person team, so there are N teams in total.\nThe tournament has a total of N-1 matches. In each match, two different teams are chosen. One team goes first, and the other goes second. Each match will result in exactly one team winning. Specifically, for each i = 1, 2, \\ldots, N-1, the i-th match proceeds as follows.\n\n- The team with player p_i goes first, and the team with player q_i goes second.\n- Let a and b be the numbers of players in the first and second teams, respectively. The first team wins with probability \\frac{a}{a+b}, and the second team wins with probability \\frac{b}{a+b}.\n- Then, the two teams are combined into a single team.\n\nThe result of each match is independent of those of the others.\nFor each of the N players, print the expected number of times the team with that player wins throughout the tournament, modulo 998244353.\n How to print an expected value modulo 998244353\nIt can be proved that the sought expected value is always rational. Also, the constraints of this problem guarantee that if the sought expected value is expressed as an irreducible fraction \\frac{y}{x}, then x is not divisible by 998244353. Now, there is a unique integer z between 0 and 998244352, inclusive, such that xz \\equiv y \\pmod{998244353}. Report this z.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\np_1 q_1\np_2 q_2\n\\vdots\np_{N-1} q_{N-1}\n\nOutput\n\nFor each i = 1, 2, \\ldots, N, print E_i, the expected number, modulo 998244353, of times the team with player i wins throughout the tournament, separated by spaces, in the following format:\nE_1 E_2 \\ldots E_N\n\nConstraints\n\n\n- 2 \\leq N \\leq 2 \\times 10^5\n- 1 \\leq p_i, q_i \\leq N\n- Just before the i-th match, player p_i and player q_i belong to different teams.\n- All input values are integers.\n\nSample Input 1\n\n5\n1 2\n4 3\n5 3\n1 4\n\nSample Output 1\n\n698771048 698771048 964969543 964969543 133099248\n\nWe call a team formed by player x_1, player x_2, \\ldots, player x_k as team \\lbrace x_1, x_2, \\ldots, x_k \\rbrace.\n\n- The first match is played by team \\lbrace 1 \\rbrace, with player 1, and team \\lbrace 2 \\rbrace, with player 2. Team \\lbrace 1 \\rbrace wins with probability \\frac{1}{2}, and team \\lbrace 2 \\rbrace wins with probability \\frac{1}{2}. Then, the two teams are combined into a single team \\lbrace 1, 2 \\rbrace.\n- The second match is played by team \\lbrace 4 \\rbrace, with player 4, and team \\lbrace 3 \\rbrace, with player 3. Team \\lbrace 4 \\rbrace wins with probability \\frac{1}{2}, and team \\lbrace 3 \\rbrace wins with probability \\frac{1}{2}. Then, the two teams are combined into a single team \\lbrace 3, 4 \\rbrace.\n- The third match is played by team \\lbrace 5 \\rbrace, with player 5, and team \\lbrace 3, 4 \\rbrace, with player 3. Team \\lbrace 5 \\rbrace wins with probability \\frac{1}{3}, and team \\lbrace 3, 4 \\rbrace wins with probability \\frac{2}{3}. Then, the two teams are combined into a single team \\lbrace 3, 4, 5 \\rbrace.\n- The fourth match is played by team \\lbrace 1, 2 \\rbrace, with player 1, and team \\lbrace 3, 4, 5 \\rbrace, with player 4. Team \\lbrace 1, 2 \\rbrace wins with probability \\frac{2}{5}, and team \\lbrace 3, 4, 5 \\rbrace wins with probability \\frac{3}{5}. Then, the two teams are combined into a single team \\lbrace 1, 2, 3, 4, 5 \\rbrace.\n\nThe expected numbers of times the teams with players 1, 2, 3, 4, 5 win throughout the tournament, E_1, E_2, E_3, E_4, E_5, are \\frac{9}{10}, \\frac{9}{10}, \\frac{53}{30}, \\frac{53}{30}, \\frac{14}{15}, respectively.\n\nSample Input 2\n\n15\n9 2\n8 10\n13 6\n12 11\n7 10\n4 10\n14 2\n5 4\n1 15\n15 2\n6 9\n8 11\n6 3\n2 8\n\nSample Output 2\n\n43970290 310168785 806914186 501498951 950708909 272140427 335124893 168750835 310168785 168750835 280459129 280459129 272140427 476542843 43970290\n\n### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.\n```python\n# YOUR CODE HERE\n```\n\n### Answer: (use the provided format with backticks)\n\nOkay, let's tackle this problem. Hmm, the problem is about calculating the expected number of wins for each player's team throughout the tournament. Each match combines two teams, and the probability of winning depends on their sizes. \n\nFirst, I need to model how the expected value accumulates for each player. Since expectation is linear, maybe I can find a way to compute for each match, how much each player contributes to their team's win probability. \n\nWait, linearity of expectation allows us to break down the total expectation into individual contributions from each match. For each player i, E_i is the sum over all matches of the probability that player i is in a team that participates in the match multiplied by their contribution to that match's probability of winning.\n\nRight. So for each match, we have two teams. Let's say during match j, the first team (a players) and second team (b players) play. Then, the first team has a/(a+b) chance to win, and the second b/(a+b). For each player in the first team, this contributes that probability to their expectation, and similarly for the second.\n\nBut how do we track which players are part of which teams during each match? That seems tricky because the matches are merging teams in a dynamic way. But perhaps there's a smarter way to model this.\n\nWait, each time two teams are merged, the structure of the tournament forms a tree-like hierarchy. Maybe we can model this using Union-Find (Disjoint Set Union) structure, but with some additional information stored for each node. \n\nAlternatively, think of each match as an edge connecting two nodes. The order of the matches determines how the teams are merged. For each player, we need to find out for every match where their team was involved, whether their team was part of the first or second team in that match, and accumulate the probabilities accordingly.\n\nBut how do we calculate the contribution for each match to each player's expectation?\n\nLet me think recursively. Let's consider that each team has a certain \"weight\" which is the number of players in it. For any player in a team, whenever their team is involved in a match, their contribution depends on the team's size at that time and the other team's size.\n\nThe challenge is to find, for each match, which players are in each of the two teams involved. Then, for each such player, add the corresponding probability (depending on which team they were in and whether their team won).\n\nBut with N up to 2e5, we need an O(N) or O(N log N) approach. So we can't process each player per match. We need to find a way to compute for each match, the contribution to all players in the team.\n\nLet me think again. Since expectation is linear, for each match, we can compute the contribution of that match to all players. For example, during the i-th match, the first team (let's call it A) and second team (B) merge. The probability that A wins is |A|/(|A|+|B|). So for all players in A, their expected value increases by |A|/(|A|+|B|), and for all players in B, it increases by |B|/(|A|+|B|). Wait no, because if A wins, then all players in A's team get +1 to their count. But since the expectation is linear, the expected value for each player in A is multiplied by the probability that their team wins in this match. So yes, for each match, each player in A contributes that probability to their expectation, same for B.\n\nSo the problem reduces to, for each match, adding to all players in team A the value of a/(a+b), and to all players in team B the value of b/(a+b). The question is how to track, for each player, all the matches in which their team participated, and which role (A or B) their team had in each match.\n\nNow, how to model this? The key is to model for each team (as it's merged over time) which matches it participates in, and in which role. However, since each team is formed by merging two teams, perhaps we can model for each node (player) the sum of probabilities contributed by each match that their team was part of.\n\nThis seems like a probl"
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