<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>

P = 1_000_000_007

def mat_pow_2x2(a, b, k):
    res1, res0 = 0, 1
    pow1, pow0 = 1, 0
    while k > 0:
        if k&1:
            c = res1*pow1
            res1, res0 = (c*a+res1*pow0+res0*pow1) % P , (c*b+res0*pow0) % P
        c = pow1*pow1
        pow1, pow0 = (c*a+2*pow1*pow0) % P , (c*b+pow0*pow0) % P
        k >>= 1
    return res1, res0

def self_matches(s):
    n = len(s)
    self_match = [0]*n
    for i in range(1, n):
        j = self_match[i-1]
        ch = s[i]
        while j > 0 and s[j] != ch:
            j = self_match[j-1]
        self_match[i] = j + int(ch==s[j])
    return self_match


class Solution:
    def numberOfWays(self, s: str, t: str, k: int) -> int:

        matches = self_matches(s+"A"+t+t)

        n = len(s)
        total = 0
        coeff1, coeff0 = mat_pow_2x2(n-2, (n-1), k)

        for i in range(2*n,3*n):
            match_len = matches[i]
            if match_len >= n:
                if i == 2*n:
                    total += coeff0
                else:
                    total += coeff1
        return total % P
1104