You are given two strings s and t of equal length n. You can perform the following operation on the string s:
s of length l where 0 < l < n and append it at the start of s.s = 'abcd' then in one operation you can remove the suffix 'cd' and append it in front of s making s = 'cdab'.You are also given an integer k. Return the number of ways in which s can be transformed into t in exactly k operations.
Since the answer can be large, return it modulo 109 + 7.
\n
Example 1:
\n\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\n\n
Example 2:
\n\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\n
\n
Constraints:
\n\n2 <= s.length <= 5 * 1051 <= k <= 1015s.length == t.lengths and t consist of only lowercase English alphabets.给你两个长度都为 n 的字符串 s 和 t 。你可以对字符串 s 执行以下操作：
s 长度为 l （0 < l < n）的 后缀字符串 删除，并将它添加在 s 的开头。s = 'abcd' ，那么一次操作中，你可以删除后缀 'cd' ，并将它添加到 s 的开头，得到 s = 'cdab' 。给你一个整数 k ，请你返回 恰好 k 次操作将 s 变为 t 的方案数。
由于答案可能很大，返回答案对 109 + 7 取余 后的结果。
\n\n
示例 1：
\n\n\n输入：s = \"abcd\", t = \"cdab\", k = 2\n输出：2\n解释：\n第一种方案：\n第一次操作，选择 index = 3 开始的后缀，得到 s = \"dabc\" 。\n第二次操作，选择 index = 3 开始的后缀，得到 s = \"cdab\" 。\n\n第二种方案：\n第一次操作，选择 index = 1 开始的后缀，得到 s = \"bcda\" 。\n第二次操作，选择 index = 1 开始的后缀，得到 s = \"cdab\" 。\n\n\n
示例 2：
\n\n\n输入：s = \"ababab\", t = \"ababab\", k = 1\n输出：2\n解释：\n第一种方案：\n选择 index = 2 开始的后缀，得到 s = \"ababab\" 。\n\n第二种方案：\n选择 index = 4 开始的后缀，得到 s = \"ababab\" 。\n\n\n
\n\n
提示：
\n\n2 <= s.length <= 5 * 1051 <= k <= 1015s.length == t.lengths 和 t 都只包含小写英文字母。t can be only constructed if it is a rotated version of string s.",
"Use KMP algorithm or Z algorithm to find the number of indices from where s is equal to t.",
"Use Dynamic Programming to count the number of ways."
],
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