<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>

def get_z(s):
    n = len(s)
    z = [0 for i in range(n)]
    best = 0
    
    for i in range(1, n):
        z[i] = max(0, min(best + z[best] - i, z[i - best]))
        while i + z[i] < n and s[i + z[i]] == s[z[i]]:
            z[i] += 1
        if best + z[best] < i + z[i]:
            best = i

    return z

MOD = 10 ** 9 + 7

def mult(a, b):
    c = [[0,0],[0,0]]
    for i in range(2):
        for j in range(2):
            for k in range(2):
                c[i][j] += a[i][k] * b[k][j]
            c[i][j] %= MOD
    return c

def pw(mat, n):
    ret = [[1, 0], [0, 1]]

    while n:
        if n & 1:
            ret = mult(ret, mat)
        mat = mult(mat, mat)
        n >>= 1
    return ret

class Solution:
    def numberOfWays(self, s: str, t: str, k: int) -> int:
        n = len(s)
        z = get_z(t + "#" + s + s)
        ways = z.count(len(t))
        ways -= int(s == t)

        if ways == 0:
            return 0

        init = [int(s != t), int(s == t)]
        g = pw([[n-ways-1, ways], [n-ways, ways-1]], k)

        return (init[0] * g[0][1] + init[1] * g[1][1]) % MOD

        
        
80000