<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:
        n = len(s)
        mod = 1000000007
        if n < 5:
            return 0
        
        dp_l = [[[0 for _ in range(11)] for _ in range(10)] for _ in range(n)]
        dp_r = [[[0 for _ in range(11)] for _ in range(10)] for _ in range(n)]
        
        for i in range(1, n):
            for j in range(10):
                for k in range(10):
                    dp_l[i][j][k] = dp_l[i-1][j][k]
            for j in range(10):
                dp_l[i][j][int(s[i-1])] += dp_l[i-1][j][10]
            for j in range(10):
                dp_l[i][j][10] = dp_l[i-1][j][10]
            dp_l[i][int(s[i-1])][10] += 1
        
        for i in range(n-2, -1, -1):
            for j in range(10):
                for k in range(10):
                    dp_r[i][j][k] = dp_r[i+1][j][k]
            for j in range(10):
                dp_r[i][int(s[i+1])][j] += dp_r[i+1][j][10]
            for j in range(10):
                dp_r[i][j][10] = dp_r[i+1][j][10]
            dp_r[i][int(s[i+1])][10] += 1
        
        ans = 0
        for i in range(2, n-2):
            for j in range(10):
                for k in range(10):
                    ans = (ans + dp_l[i][j][k]*dp_r[i][k][j]%mod) % mod
        
        return ans
64800