<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:
    MOD = 10 ** 9 + 7
    P = 31
    PMOD = 496822652529863993

    def numberOfWays(self, s: str, t: str, k: int) -> int:
        n = len(s)

        sHash = 0
        for c in s:
            sHash *= Solution.P
            sHash += ord(c) - ord('a')
            sHash %= Solution.PMOD

        tHash = 0
        for c in t:
            tHash *= Solution.P
            tHash += ord(c) - ord('a')
            tHash %= Solution.PMOD

        # Rabin-Karp
        res = 0
        largestPower = pow(Solution.P, n - 1, Solution.PMOD)
        fk1 = (pow((n - 1), k, Solution.MOD) - (-1) ** k) * pow(n, -1, Solution.MOD)  # f(k, 1)
        for i, c in enumerate(t):
            # cyclic substring found
            if sHash == tHash:
                res += fk1 if i != 0 else fk1 + (-1) ** k
                res %= Solution.MOD

            # shift t
            intC = ord(c) - ord('a')
            tHash -= largestPower * intC
            tHash = (tHash % Solution.PMOD + Solution.PMOD) % Solution.PMOD

            tHash *= Solution.P
            tHash += intC
            tHash %= Solution.PMOD

        return res
1217