<p>Given a string s, return <em>the number of different non-empty palindromic subsequences in</em> <code>s</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>A subsequence of a string is obtained by deleting zero or more characters from the string.</p>

<p>A sequence is palindromic if it is equal to the sequence reversed.</p>

<p>Two sequences <code>a<sub>1</sub>, a<sub>2</sub>, ...</code> and <code>b<sub>1</sub>, b<sub>2</sub>, ...</code> are different if there is some <code>i</code> for which <code>a<sub>i</sub> != b<sub>i</sub></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bccb&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> The 6 different non-empty palindromic subsequences are &#39;b&#39;, &#39;c&#39;, &#39;bb&#39;, &#39;cc&#39;, &#39;bcb&#39;, &#39;bccb&#39;.
Note that &#39;bcb&#39; is counted only once, even though it occurs twice.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba&quot;
<strong>Output:</strong> 104860361
<strong>Explanation:</strong> There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10<sup>9</sup> + 7.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s[i]</code> is either <code>&#39;a&#39;</code>, <code>&#39;b&#39;</code>, <code>&#39;c&#39;</code>, or <code>&#39;d&#39;</code>.</li>
</ul>

class Solution:
    def countPalindromicSubsequences(self, s: str) -> int:
        N = len(s)
        MOD = 10**9 + 7
        memo = dict()

        def impl(i, j):
            if i > j:
                return 0
            elif i == j:
                return 1
            elif i + 1 == j:
                return 2
            if (i, j) in memo:
                return memo[(i, j)]
            if s[i] == s[j]:
                res = 2 * impl(i + 1, j - 1)
                res %= MOD
                left = i + 1
                right = j - 1
                while left <= right:
                    if s[left] == s[i]:
                        break
                    left += 1
                while left <= right:
                    if s[right] == s[i]:
                        break
                    right -= 1
                if left < right:
                    res += MOD - impl(left + 1, right - 1)
                    res %= MOD
                elif left == right:
                    res += 1
                    res %= MOD
                else:
                    res += 2
                    res %= MOD
            else:
                res = impl(i + 1, j) % MOD
                res += impl(i, j - 1)
                res %= MOD
                res += MOD - impl(i + 1, j - 1)
                res %= MOD
            memo[(i, j)] = res
            return res

        res = impl(0, N - 1)
        return res

168729