<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:
        dp = [{},{},{},{}]
        ans = 0
        inf = 10**9+7
        for c in s:
            for k in dp[3].keys():
                if c==k[0]: ans += dp[3][k]
                if ans >= inf: ans %= inf
            for k in dp[2].keys():
                if c==k[1]: dp[3][k+c] = dp[3].get(k+c,0)+dp[2][k]
            for i in range(2,0,-1):
                for k in dp[i-1].keys():
                    dp[i][k+c] = dp[i].get(k+c,0)+dp[i-1][k]
            dp[0][c] = dp[0].get(c,0)+1
        return ans
17000