<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:
        # 3 1 2 3 5 6 3 4 5 3
        N = len(s)
        MOD = int(1e9 + 7)

        # a b c b a
        num_to_count = [0] * 10
        idx_to_count = []
        for i in range(N):
            temp = [[0] * 10 for _ in range(10)]
            idx_to_count.append(temp)

        for i in range(N):
            x = ord(s[i]) - ord('0')

            if i > 0:
                for a in range(10):
                    for b in range(10):
                        idx_to_count[i][a][b] = idx_to_count[i - 1][a][b]
                    idx_to_count[i][a][x] += num_to_count[a]

            num_to_count[x] += 1

        num_to_count = [0] * 10
        count = [[0] * 10 for _ in range(10)]

        total = 0
        for i in range(N - 1, -1, -1):
            if i <= 1:
                break
            x = ord(s[i]) - ord('0')
            for a in range(10):
                for b in range(10):
                    total += idx_to_count[i - 1][a][b] * count[b][a]
                    total %= MOD
            
            for a in range(10):
                count[x][a] += num_to_count[a]
            num_to_count[x] += 1

        return total
35600