<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>

from functools import lru_cache

class Solution:
    def findGoodStrings(self, n: int, s1: str, s2: str, evil: str) -> int:
        A1=[ord(c)-ord('a') for c in s1]
        A2=[ord(c)-ord('a') for c in s2]
        
        mod=10**9+7
        def KMP(s):
            n=len(s)
            p=[0]*n
            for i in range(1,n):
                j=p[i-1]
                while j and s[i]!=s[j]:
                    j=p[j-1]
                if s[i]==s[j]:j+=1
                p[i]=j
            return p
        
        kmp=KMP(evil)
        x=[ord(c)-ord('a') for c in evil]
        
        @lru_cache(None)
        def dp(i,j,flag):  
            if j==len(evil):
                return 0
            if i==n:
                return 1
            
            ans=0
            for ci in (range(A2[i]+1) if flag else range(26)):
                j2=j
                while j2 and ci!=x[j2]:
                    j2=kmp[j2-1]

                if ci==x[j2]:
                    j2+=1
                flag2=flag and ci==A2[i]
                ans+=dp(i+1,j2,flag2)
                ans%=mod
            return ans
        
        @lru_cache(None)
        def dp1(i,j,flag):  
            if j==len(evil):
                return 0
            if i==n:
                return 1 if not flag else 0
            
            ans=0
            for ci in (range(A1[i]+1) if flag else range(26)):
                j2=j
                while j2 and ci!=x[j2]:
                    j2=kmp[j2-1]

                if ci==x[j2]:
                    j2+=1
                
                flag2=flag and ci==A1[i]
                ans += dp1(i+1,j2,flag2)
                ans%=mod
            return ans
        
        top=dp(0,0,True)
        bot=dp1(0,0,True)
        return (top-bot)%mod
24200