<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 kmp(self, s, t):
        pi, res = [0 for i in range(len(t))], []
        for i in range(1, len(t)):
            j = pi[i-1]
            while j > 0 and t[j] != t[i]: j = pi[j-1]
            pi[i] = 0 if j == 0 and t[0] != t[i] else j + 1
        m, n, j = len(s), len(t), 0
        for i in range(m):
            while j >= n or j > 0 and s[i] != t[j]: j = pi[j-1]
            if s[i] == t[j]: j += 1
            if j == n: res.append(i - n + 1)
        return res
    def numberOfWays(self, s: str, t: str, k: int) -> int:
        n, M = len(s), 10**9 + 7
        pos = self.kmp((s+s)[:-1], t)
        f_k = [0, 0]
        f_k[1] = (pow(n-1, k, M) - (-1)**k + M) * pow(n, M-2, M) % M
        f_k[0] = (f_k[1] + (-1)**k + M) % M
        return sum(f_k[not not p] for p in pos) % M
        #too complicated. didn't understand
607