<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>

import numpy as np

class Solution:
    def numberOfWays(self, s: str, t: str, k: int) -> int:
        def matrix_pow(m, k):
            if k == 1:
                return m
            h = matrix_pow(m, k//2)
            if k % 2 == 0:
                return np.matmul(h, h) % 1000000007
            else:
                return np.matmul(np.matmul(h, h) % 1000000007, m) % 1000000007

        n = len(s)
        P = 31
        M = 1000000009+9
        p_pow_n = pow(P, n, M)
        ht = 0
        ord_a = ord('a')
        for c in t:
            ht = (ht * P + (ord(c) - ord_a + 1)) % M
        targets = 0
        hs = 0
        for i, c in enumerate(s+s):
            hs = (hs * P + (ord(c) - ord_a + 1)) % M
            if i >= n:
                hs = (hs - (ord(s[i-n]) - ord_a + 1) * p_pow_n) % M
                if ht == hs:
                    targets += 1
        if targets == 0:
            return 0
        src = 1 if t == s[:n] else 0
        m = matrix_pow(np.array([[n - targets - 1, targets], [n - targets, targets - 1]]), k)
        return m[src][1]
968