<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 prefix_count(self, arr):
        n = len(arr)
        assert n >= 2

        prefix_cnt = [[0] * 10 for _ in range(n)]
        prefix_cnt[0][arr[0]] = 1
        for i in range(1, n):
            for j in range(10):
                if arr[i] == j:
                    prefix_cnt[i][j] = prefix_cnt[i-1][j] + 1
                else:
                    prefix_cnt[i][j] = prefix_cnt[i-1][j]

        cnt = [[0] * 100 for _ in range(n)]
        for i in range(1, n):
            for j in range(100):
                if arr[i] == j % 10:
                    cnt[i][j] = prefix_cnt[i-1][j // 10]
        
        result = [[0] * 100 for _ in range(n)]
        for i in range(1, n):
            for j in range(100):
                result[i][j] = result[i-1][j] + cnt[i][j]

        return result

    def countPalindromes(self, s: str) -> int:
        n = len(s)
        if n < 5:
            return 0

        arr = [int(c) for c in s]
        pc = self.prefix_count(arr)
        pc_rev = self.prefix_count(arr[::-1])

        total = 0
        for i in range(2, n-2):
            for j in range(100):
                total = (total + pc[i-1][j] * pc_rev[n-i-2][j]) % (int(1e9) + 7)

        return total
105600