<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 numberOfWays(self, s: str, t: str, k: int) -> int:
        M = 10 ** 9 + 7
        cnt = Counter(s)
        n = len(s)
        i = (s + s).find(t)
        if i == -1:
            return 0
        if len(cnt) == 1:
            return pow(n - 1, k, M) % M
        elif len(cnt) == 2:
            if s == s[:2] * (n // 2):
                res = n // 2
                if s == t:
                    curr = pow(n - 1, k, M) * res
                    if k & 1:
                        curr -= n - res
                    else:
                        curr += n - res
                    return curr * pow(n, -1, M) % M
                else:
                    curr = pow(n - 1, k, M) * res
                    if k & 1:
                        curr += res
                    else:
                        curr -= res
                    return curr * pow(n, -1, M) % M
            


        ss = s[i+1:] + s
        res = 1
        while len(ss) > n:
            i = ss.find(t)
            if i == -1:
                break
            if i <= len(ss) - n - 1:
                res += 1
            ss = ss[i+1:]
        
        if s == t:
            curr = pow(n - 1, k, M) * res
            if k & 1:
                curr -= n - res
            else:
                curr += n - res
            return curr * pow(n, -1, M) % M
        else:
            curr = pow(n - 1, k, M) * res
            if k & 1:
                curr += res
            else:
                curr -= res
            return curr * pow(n, -1, M) % M
810