<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, MOD, n = 10, int(1e9) + 7, len(s)
        prefix = [[[0] * N for _ in range(N)] for _ in range(n)]
        count = [0] * N

        L = tuple(enumerate(map(ord, s)))

        for i, c in L:
            c -= 48
            if i:
                for j in range(N):
                    for k in range(N):
                        prefix[i][j][k] = prefix[i - 1][j][k]
                        if k == c: prefix[i][j][k] += count[j]
            count[c] += 1

        suffix = [[[0] * N for _ in range(N)] for _ in range(len(s))]
        count = [0] * N

        for i, c in reversed(L):
            c -= 48
            if i + 1 < n:
                for j in range(N):
                    for k in range(N):
                        suffix[i][j][k] = suffix[i + 1][j][k]
                        if k == c: suffix[i][j][k] += count[j]
            count[c] += 1

        res = 0
        for i in range(2, n - 2):
            for j in range(N):
                for k in range(N):
                    res += prefix[i - 1][j][k] * suffix[i + 1][j][k] % MOD

        return res % MOD
55800