<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:
        MOD = 10**9 + 7

        # KMP Automaton
        def kmp_automaton(s):
            m = len(s)
            automaton = [0] * (m + 1)
            j = 0
            for i in range(1, m):
                while j > 0 and s[i] != s[j]:
                    j = automaton[j - 1]
                if s[i] == s[j]:
                    j += 1
                automaton[i] = j
            return automaton

        # DP Transitions
        def dp(s1, s2, evil):
            m = len(evil)
            automaton = kmp_automaton(evil)
            
            @functools.lru_cache(None)
            def dfs(pos, tight1, tight2, state):
                if state == m:
                    return 0
                if pos == n:
                    return 1
                
                start = ord(s1[pos]) if tight1 else ord('a')
                end = ord(s2[pos]) if tight2 else ord('z')
                
                count = 0
                for ch in range(start, end + 1):
                    new_tight1 = tight1 and (ch == start)
                    new_tight2 = tight2 and (ch == end)
                    new_state = state
                    while new_state > 0 and chr(ch) != evil[new_state]:
                        new_state = automaton[new_state - 1]
                    if chr(ch) == evil[new_state]:
                        new_state += 1
                    count += dfs(pos + 1, new_tight1, new_tight2, new_state)
                    count %= MOD
                
                return count
            
            return dfs(0, True, True, 0)
        
        return dp(s1, s2, evil)

# Example usage:
solution = Solution()
n = 2
s1 = "aa"
s2 = "da"
evil = "b"
print(solution.findGoodStrings(n, s1, s2, evil))  # Output: 51
22200