<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:
        mod = 10**9+7
        dp = [[[0,0] for _ in range(100)] for __ in range(len(s))]
        digits_before = [0] * 10
        for i in range(len(s)-1,-1,-1):
            cur_digit = ord(s[i]) - ord('0')
            for j in range(100):
                if i < len(s)-1:
                    dp[i][j][0] = dp[i+1][j][0]
                    dp[i][j][1] = dp[i+1][j][0] + dp[i+1][j][1]
            for d,c in enumerate(digits_before):
                cur_pair = cur_digit*10 + d
                dp[i][cur_pair][0] += c
            digits_before[cur_digit]+=1
        digits_before = [0] * 10
        res = 0
        for i in range(len(s)-1):
            for d in range(10):
                num_left_pairs = digits_before[d]
                right_pair = (ord(s[i]) - ord('0')) * 10 + d
                right_distance = dp[i+1][right_pair][1]
                res += right_distance * num_left_pairs
                res %= mod
            digits_before[ord(s[i]) - ord('0')] += 1
        return res

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