<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:
		letter_count_left = {}
		pair_count_left = {}
		letter_count_right = {}
		pair_count_right = {}
		N = len(s)
		
		for i in range(N):
			pair_count_left[i] = pair_count_left[i - 1].copy() if i - 1 >= 0 else {}
			curr = s[i]
			for prev, cnt in letter_count_left.items():
				key = prev + curr
				if key in pair_count_left[i]:
					pair_count_left[i][key] += cnt 
				else:
					pair_count_left[i][key] = cnt 
			if curr in letter_count_left:
				letter_count_left[curr] += 1
			else:
				letter_count_left[curr] = 1
				
			j = N - 1 - i
			pair_count_right[j] = pair_count_right[j + 1].copy() if j + 1 < N else {}
			curr = s[j]
			for prev, cnt in letter_count_right.items():
				key = prev + curr
				if key in pair_count_right[j]:
					pair_count_right[j][key] += cnt 
				else:
					pair_count_right[j][key] = cnt 
			
			if curr in letter_count_right:
				letter_count_right[curr] += 1
			else:
				letter_count_right[curr] = 1
		
		result = 0
		mod = 10**9 + 7
		for i in range(2, N - 2):
			cand_left = pair_count_left[i - 1]
			cand_right = pair_count_right[i + 1]
			for pair1 in cand_left:
				for pair2 in cand_right:
					if pair1 == pair2:
						result += cand_left[pair1] * cand_right[pair2]
		return result % mod 
89000