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Given a string s, we make queries on substrings of s.
For each query queries[i] = [left, right, k], we may rearrange the substring s[left], ..., s[right], and then choose up to k of them to replace with any lowercase English letter.
If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false.
Return an array answer[], where answer[i] is the result of the i-th query queries[i].
Note that: Each letter is counted individually for replacement so if for example s[left..right] = "aaa", and k = 2, we can only replace two of the letters. (Also, note that the initial string s is never modified by any query.)
Example :
Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]] Output: [true,false,false,true,true] Explanation: queries[0] : substring = "d", is palidrome. queries[1] : substring = "bc", is not palidrome. queries[2] : substring = "abcd", is not palidrome after replacing only 1 character. queries[3] : substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab". queries[4] : substring = "abcda", could be changed to "abcba" which is palidrome.
Constraints:
1 <= s.length, queries.length <= 10^50 <= queries[i][0] <= queries[i][1] < s.length0 <= queries[i][2] <= s.lengthsonly contains lowercase English letters.
给你一个字符串 s，请你对 s 的子串进行检测。
每次检测，待检子串都可以表示为 queries[i] = [left, right, k]。我们可以 重新排列 子串 s[left], ..., s[right]，并从中选择 最多 k 项替换成任何小写英文字母。
如果在上述检测过程中，子串可以变成回文形式的字符串，那么检测结果为 true，否则结果为 false。
返回答案数组 answer[]，其中 answer[i] 是第 i 个待检子串 queries[i] 的检测结果。
注意：在替换时，子串中的每个字母都必须作为 独立的 项进行计数，也就是说，如果 s[left..right] = "aaa" 且 k = 2，我们只能替换其中的两个字母。（另外，任何检测都不会修改原始字符串 s，可以认为每次检测都是独立的）
示例：
输入：s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]] 输出：[true,false,false,true,true] 解释： queries[0] : 子串 = "d"，回文。 queries[1] : 子串 = "bc"，不是回文。 queries[2] : 子串 = "abcd"，只替换 1 个字符是变不成回文串的。 queries[3] : 子串 = "abcd"，可以变成回文的 "abba"。 也可以变成 "baab"，先重新排序变成 "bacd"，然后把 "cd" 替换为 "ab"。 queries[4] : 子串 = "abcda"，可以变成回文的 "abcba"。
提示：
1 <= s.length, queries.length <= 10^50 <= queries[i][0] <= queries[i][1] < s.length0 <= queries[i][2] <= s.lengths中只有小写英文字母