<p>Given the strings <code>s1</code> and <code>s2</code> of size <code>n</code> and the string <code>evil</code>, return <em>the number of <strong>good</strong> strings</em>.</p>

<p>A <strong>good</strong> string has size <code>n</code>, it is alphabetically greater than or equal to <code>s1</code>, it is alphabetically smaller than or equal to <code>s2</code>, and it does not contain the string <code>evil</code> as a substring. Since the answer can be a huge number, return this <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> n = 2, s1 = &quot;aa&quot;, s2 = &quot;da&quot;, evil = &quot;b&quot;
<strong>Output:</strong> 51 
<strong>Explanation:</strong> There are 25 good strings starting with &#39;a&#39;: &quot;aa&quot;,&quot;ac&quot;,&quot;ad&quot;,...,&quot;az&quot;. Then there are 25 good strings starting with &#39;c&#39;: &quot;ca&quot;,&quot;cc&quot;,&quot;cd&quot;,...,&quot;cz&quot; and finally there is one good string starting with &#39;d&#39;: &quot;da&quot;.&nbsp;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 8, s1 = &quot;leetcode&quot;, s2 = &quot;leetgoes&quot;, evil = &quot;leet&quot;
<strong>Output:</strong> 0 
<strong>Explanation:</strong> All strings greater than or equal to s1 and smaller than or equal to s2 start with the prefix &quot;leet&quot;, therefore, there is not any good string.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 2, s1 = &quot;gx&quot;, s2 = &quot;gz&quot;, evil = &quot;x&quot;
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>s1.length == n</code></li>
	<li><code>s2.length == n</code></li>
	<li><code>s1 &lt;= s2</code></li>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>1 &lt;= evil.length &lt;= 50</code></li>
	<li>All strings consist of lowercase English letters.</li>
</ul>

class Solution:
    def findGoodStrings(self, n: int, s1: str, s2: str, evil: str) -> int:
        chars, lenEvil = string.ascii_lowercase, len(evil)
        lps, left = [0] * lenEvil, 0
        for right in range(1, lenEvil):
            while left and evil[left] != evil[right]:
                left = lps[left - 1]
            if evil[left] == evil[right]:
                left += 1
            lps[right] = left
        
        @cache
        def dp(charIdx: int, largerThanS1: bool, smallerThanS2: bool, evilIdx: int) -> int:
            if charIdx == n:
                return 1
            ways = 0
            for char in chars:
                if not largerThanS1 and char < s1[charIdx]: continue
                if not smallerThanS2 and char > s2[charIdx]: continue
                if evilIdx == lenEvil - 1 and char == evil[evilIdx]: continue
                if char == evil[evilIdx]:
                    newEvilIdx = evilIdx + 1
                else:
                    newEvilIdx = evilIdx
                    while newEvilIdx and evil[newEvilIdx] != char:
                        newEvilIdx = lps[newEvilIdx - 1]
                    if evil[newEvilIdx] == char:
                        newEvilIdx += 1
                ways += dp(charIdx + 1, largerThanS1 | (char > s1[charIdx]), smallerThanS2 | (char < s2[charIdx]), newEvilIdx)
            return ways % 1000000007

        return dp(0, False, False, 0)
21600