<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 {
public:
    #define ll long long
    ll mod = 1e9+7;
    ll dp[1001][1001][5];
    ll f(ll i,ll j,ll ch, string &s){
        if(i>j)return 0;
        if(i==j){
            return (s[i]==('a'+ch));
        }
        if(dp[i][j][ch]!=-1)return dp[i][j][ch];
        ll cnt=0;
        if((s[i]==s[j]) && (s[i]==(char)(ch+'a'))){
            cnt=2;
            for(ll x=0;x<4;x++){
                cnt = (cnt + f(i+1,j-1,x,s))%mod;
            }
        }else{
            cnt = (cnt + f(i+1,j,ch,s) + f(i,j-1,ch,s)-f(i+1,j-1,ch,s))%mod;
        }
        return dp[i][j][ch]=cnt;
    }
    int countPalindromicSubsequences(string s) {
        ll n = s.size();
        memset(dp,-1,sizeof(dp));
        ll ans = 0;
        for(int i=0;i<4;i++){
            ans = (ans+f(0,n-1,i,s))%mod;
        }
        return ans;
    }
};
47151