<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:
        L = len(s)
        
        pre = [[[0]* 10 for _ in range(10)] for _ in range(L)]
        suf = [[[0]* 10 for _ in range(10)] for _ in range(L)]
        
        counts = [0] * 10
        counts[int(s[0])] = 1
        for idx in range(1, L):
            num = int(s[idx])
            for num1 in range(10):
                for num2 in range(10):
                    # if num2== 0:
                    #     print(idx, 'pre[idx-1][num1][num2]',pre[idx-1][num1][num2])
                    pre[idx][num1][num2] = pre[idx-1][num1][num2]
                
                    if num2 == num:

                        pre[idx][num1][num2] += counts[num1]
            counts[num]+=1
        # print('pre',pre[L-1][1][0])
        
        counts = [0] * 10
        counts[int(s[-1])] = 1
        # print('czxc',counts)
        for idx in range(L-2, -1,-1):
            num = int(s[idx])
            for num1 in range(10):
                for num2 in range(10):
                    # if num2== 0:
                        # print(idx,num2,num1, 'suf[idx-1][num1][num2]',suf[idx+1][num1][num2])
                    suf[idx][num2][num1] = suf[idx+1][num2][num1]
                
                    if num2 == num:
                        suf[idx][num2][num1] += counts[num1]
                        # print('zz',)
            counts[num]+=1
        # print('suffix',suf[0][1][0])
        ans = 0 
        # for i in range(0, L-1):
            # print('das',suf[i][0][1], suf[i][1][0])
        for i in range(2,L-2):
            for j in range(10):
                for k in range(10):
                    # print('i,j,k', i,j,k)
                    # if j == 1 and k==0:
                        # print('asdfds',i,pre[i-1][j][k], suf[i+1][k][j])
                    ans += pre[i-1][j][k] * suf[i+1][k][j]
        return ans %(10**9+7)
        # Time O(N*100) = O(N)
        # Spae O(N*100) = O(N)
        # pre = []
        # 10 ... 01
        # pre[1][0] = 1
        # suf[0][1]
54700