<p>You are given two strings <code>s</code> and <code>t</code> of equal length <code>n</code>. You can perform the following operation on the string <code>s</code>:</p>

<ul>
	<li>Remove a <strong>suffix</strong> of <code>s</code> of length <code>l</code> where <code>0 &lt; l &lt; n</code> and append it at the start of <code>s</code>.<br />
	For example, let <code>s = &#39;abcd&#39;</code> then in one operation you can remove the suffix <code>&#39;cd&#39;</code> and append it in front of <code>s</code> making <code>s = &#39;cdab&#39;</code>.</li>
</ul>

<p>You are also given an integer <code>k</code>. Return <em>the number of ways in which </em><code>s</code> <em>can be transformed into </em><code>t</code><em> in <strong>exactly</strong> </em><code>k</code><em> operations.</em></p>

<p>Since the answer can be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcd&quot;, t = &quot;cdab&quot;, k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
First way:
In first operation, choose suffix from index = 3, so resulting s = &quot;dabc&quot;.
In second operation, choose suffix from index = 3, so resulting s = &quot;cdab&quot;.

Second way:
In first operation, choose suffix from index = 1, so resulting s = &quot;bcda&quot;.
In second operation, choose suffix from index = 1, so resulting s = &quot;cdab&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ababab&quot;, t = &quot;ababab&quot;, k = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
First way:
Choose suffix from index = 2, so resulting s = &quot;ababab&quot;.

Second way:
Choose suffix from index = 4, so resulting s = &quot;ababab&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 5 * 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>15</sup></code></li>
	<li><code>s.length == t.length</code></li>
	<li><code>s</code> and <code>t</code> consist of only lowercase English alphabets.</li>
</ul>

class Solution:
    def compute_lps_array(self, pattern):
        length = 0
        lps = [0] * len(pattern)
        i = 1

        while i < len(pattern):
            if pattern[i] == pattern[length]:
                length += 1
                lps[i] = length
                i += 1
            else:
                if length != 0:
                    length = lps[length - 1]
                else:
                    lps[i] = 0
                    i += 1
        return lps

    def KMP_search(self, text, pattern):
        m = len(pattern)
        n = len(text)
        lps = self.compute_lps_array(pattern)
        cnt = 0
        i = j = 0
        while i < n:
            if pattern[j] == text[i]:
                i += 1
                j += 1

            if j == m:
                cnt += 1
                j = lps[j - 1]
            elif i < n and pattern[j] != text[i]:
                if j != 0:
                    j = lps[j - 1]
                else:
                    i += 1
        return cnt

    def numberOfWays(self, A: str, B: str, k: int) -> int:
        MOD = 10**9 + 7
        ka, kb = A == B, self.KMP_search(A[1:] + A[:-1], B)
        n = len(A)
        f = (pow(n - 1, k, MOD) - pow(-1, k)) * pow(n, MOD - 2, MOD)
        return ((ka+kb)*f+ka*pow(-1,k))%MOD
87