<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 
{
public:
    int dp[501][51][2][2], n, m;
    vector<int> kmp;
    int findGoodStrings(int N, string &s1, string &s2, string &evil) 
    {
        n = N;
        kmp = KMP(evil);
        memset(dp, -1, sizeof(dp));
        return dfs(0, 0, 1, 1, s1, s2, evil);
    }
private:
    vector<int> KMP(string &s)
    {
        m = s.size();
        int k = 0;
        vector<int> kmp(m, 0);
        for (int i = 1; i < m; i++)
        {
            while (k>0 && s[i] != s[k]) k = kmp[k-1];
            k = kmp[i] = k + (s[i] == s[k]);
        }
        return kmp;
    }
    int dfs(int i, int spE, bool sameS1, bool sameS2, string &s1, string &s2, string &evil)
    {
        if (spE == m) return 0;
        if (i == n) return 1;
        if (dp[i][spE][sameS1][sameS2] != -1) return dp[i][spE][sameS1][sameS2];
        int res = 0;
        for (char j = (sameS1 ? s1[i] : 'a'); j <= (sameS2 ? s2[i] : 'z'); j++)
        {
            int tmp = spE;
            while (tmp > 0 && j != evil[tmp]) tmp = kmp[tmp-1];
            tmp = tmp + (evil[tmp] == j);
            res += dfs(i+1, tmp, sameS1 && (j == s1[i]), sameS2 && (j == s2[i]), s1, s2, evil);
            res %= (int)(1e9 + 7);
        }
        return dp[i][spE][sameS1][sameS2] = res;
    }
};
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