<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:
        MOD = 10**9 + 7
        n = len(s)
        dp = [[[0 for _ in range(4)] for _ in range(n)] for _ in range(n)]

        for i in range(n):
            for k in range(4):
                if s[i] == chr(ord('a') + k):
                    dp[i][i][k] = 1

        for length in range(2, n + 1):
            for i in range(n - length + 1):
                j = i + length - 1
                for k in range(4):
                    c = chr(ord('a') + k)
                    if s[i] != c:
                        dp[i][j][k] = dp[i+1][j][k]
                    elif s[j] != c:
                        dp[i][j][k] = dp[i][j-1][k]
                    else:
                        # s[i] == s[j] == c
                        dp[i][j][k] = 2
                        if length > 2:
                            for l in range(4):
                                dp[i][j][k] += dp[i+1][j-1][l]
                                dp[i][j][k] %= MOD

        result = 0
        for k in range(4):
            result += dp[0][n-1][k]
            result %= MOD

        return result

# Example usage:
solution = Solution()
print(solution.countPalindromicSubsequences("bccb"))  # Output: 6
print(solution.countPalindromicSubsequences("abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba"))  # Output: 104860361
112231