<p>Given a string of digits <code>s</code>, return <em>the number of <strong>palindromic subsequences</strong> of</em> <code>s</code><em> having length </em><code>5</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p><strong>Note:</strong></p>

<ul>
	<li>A string is <strong>palindromic</strong> if it reads the same forward and backward.</li>
	<li>A <strong>subsequence</strong> is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.</li>
</ul>

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

<pre>
<strong>Input:</strong> s = &quot;103301&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
There are 6 possible subsequences of length 5: &quot;10330&quot;,&quot;10331&quot;,&quot;10301&quot;,&quot;10301&quot;,&quot;13301&quot;,&quot;03301&quot;. 
Two of them (both equal to &quot;10301&quot;) are palindromic.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;0000000&quot;
<strong>Output:</strong> 21
<strong>Explanation:</strong> All 21 subsequences are &quot;00000&quot;, which is palindromic.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;9999900000&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The only two palindromic subsequences are &quot;99999&quot; and &quot;00000&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of digits.</li>
</ul>

class Solution:
    def countPalindromes(self, s: str) -> int:
        '''
        f(j,k) = f(j-1,k) + f(j-1,k-1)*int(s[j]==p[k])
        '''
        ans = 0
        n = len(s)
        for i in range(100):
            p = str(i%10) + str((i//10)%10)
            cl, cr = self.helper(s,p), self.helper(s[::-1],p)
            for i in range(2,n-2):
                ans += cl[i-1]*cr[n-i-2]
        return ans % 1000000007

    def helper(self, s, p):
        n = len(s)
        m = len(p)
        mtx = [[0 for i in range(n)] for j in range(m)]
        mtx[0][0] = int(p[0]==s[0])
        for i in range(1,n):
            mtx[0][i] = mtx[0][i-1] + int(p[0]==s[i])
        for j in range(1,m):
            for i in range(j,n):
                mtx[j][i] = mtx[j][i-1] + mtx[j-1][i-1]*int(p[j]==s[i])
        return mtx[-1]
18500