<p>Given a string s, return <em>the number of different non-empty palindromic subsequences in</em> <code>s</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>A subsequence of a string is obtained by deleting zero or more characters from the string.</p>

<p>A sequence is palindromic if it is equal to the sequence reversed.</p>

<p>Two sequences <code>a<sub>1</sub>, a<sub>2</sub>, ...</code> and <code>b<sub>1</sub>, b<sub>2</sub>, ...</code> are different if there is some <code>i</code> for which <code>a<sub>i</sub> != b<sub>i</sub></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bccb&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> The 6 different non-empty palindromic subsequences are &#39;b&#39;, &#39;c&#39;, &#39;bb&#39;, &#39;cc&#39;, &#39;bcb&#39;, &#39;bccb&#39;.
Note that &#39;bcb&#39; is counted only once, even though it occurs twice.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba&quot;
<strong>Output:</strong> 104860361
<strong>Explanation:</strong> There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10<sup>9</sup> + 7.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s[i]</code> is either <code>&#39;a&#39;</code>, <code>&#39;b&#39;</code>, <code>&#39;c&#39;</code>, or <code>&#39;d&#39;</code>.</li>
</ul>

class Solution:
    def recurse(self,i,j,s,c,dp):
        if i>j:return 0
        if i==j:
            if s[i]==c:return 1
            return 0
        if dp[i][j][ord(c)-ord('a')]!=-1:
            return dp[i][j][ord(c)-ord('a')]
        count=0
        if s[i]==s[j] and s[i]==c:
            count+=2
            for ch in ['a','b','c','d']:
                count=(count+self.recurse(i+1,j-1,s,ch,dp))%self.mod
        else:
            count=(count+self.recurse(i+1,j,s,c,dp))%self.mod
            count=(count+self.recurse(i,j-1,s,c,dp))%self.mod
            count=(count-self.recurse(i+1,j-1,s,c,dp))%self.mod
            count=count if count>=0 else count+self.mod
        dp[i][j][ord(c)-ord('a')]=count
        return count
            
        
        
    def countPalindromicSubsequences(self, s: str) -> int:
        self.mod=1000_000_007
        dp=[[[-1 for _ in range(4)] for _ in range(len(s)+1)]for _ in range(len(s)+1)]
        ans=0
        for i in ['a','b','c','d']:
            ans=(ans+self.recurse(0,len(s)-1,s,i,dp))%self.mod
        return ans
        
140480