<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>

from collections import Counter
class Solution:
    def countPalindromes(self, s: str) -> int:
        '''
        mod, n = int(10**9)+7, len(s)
        dp = [[0 for _ in range(n)] for _ in range(n)]
        for i in range(n-2, -1, -1):
            for j in range(i+2, n):
                dp[i][j] = dp[i][j-1] + (0 if dp[i + 1][j] == dp[i + 1][j-1] else dp[i + 1][j] - dp[i + 1][j - 1])
                if s[i] == s[j]:
                    dp[i][j] += j - i - 1
        res = 0
        for i in range(n):
            for j in range(i+4, n):
                if s[i] == s[j]:
                    res += dp[i+1][j-1]
                    res %= mod
        return res
        '''
            
        n = len(s)
        p1, s1 = [[0 for _ in range(10)] for _ in range(n+1)], [[0 for _ in range(10)] for _ in range(n+1)]
        prefix, suffix = [Counter() for _ in range(n)], [Counter() for _ in range(n)]
        mod, res = int(10**9)+7, 0
        for i in range(n):
            if i > 0:
                prefix[i] = Counter(prefix[i-1])
            for j in range(10):
                p = str(j)+s[i]
                prefix[i][p] += p1[i][j]
            for j in range(10):
                p1[i+1][j] = p1[i][j] + (1 if int(s[i])==j else 0)
        for i in range(n-1, -1, -1):
            if i < n-1:
                suffix[i] = Counter(suffix[i+1])
            for j in range(10):
                temp = str(j)+s[i]
                suffix[i][temp] += s1[i+1][j]
            if 2 <= i < n - 1:
                for p in suffix[i]:
                    res += prefix[i-2][p]*suffix[i][p]
                    res %= mod
            for j in range(10):
                s1[i][j] = s1[i+1][j] + (1 if int(s[i])==j else 0)
  
        return res
97200