<p>You are given a string <code>s</code> of length <code>n</code> where <code>s[i]</code> is either:</p>

<ul>
	<li><code>&#39;D&#39;</code> means decreasing, or</li>
	<li><code>&#39;I&#39;</code> means increasing.</li>
</ul>

<p>A permutation <code>perm</code> of <code>n + 1</code> integers of all the integers in the range <code>[0, n]</code> is called a <strong>valid permutation</strong> if for all valid <code>i</code>:</p>

<ul>
	<li>If <code>s[i] == &#39;D&#39;</code>, then <code>perm[i] &gt; perm[i + 1]</code>, and</li>
	<li>If <code>s[i] == &#39;I&#39;</code>, then <code>perm[i] &lt; perm[i + 1]</code>.</li>
</ul>

<p>Return <em>the number of <strong>valid permutations</strong> </em><code>perm</code>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;DID&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> The 5 valid permutations of (0, 1, 2, 3) are:
(1, 0, 3, 2)
(2, 0, 3, 1)
(2, 1, 3, 0)
(3, 0, 2, 1)
(3, 1, 2, 0)
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;D&quot;
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>n == s.length</code></li>
	<li><code>1 &lt;= n &lt;= 200</code></li>
	<li><code>s[i]</code> is either <code>&#39;I&#39;</code> or <code>&#39;D&#39;</code>.</li>
</ul>

class Solution:
    def numPermsDISequence(self, s: str) -> int:
        MOD = 10**9 + 7
        n = len(s)
        # dp[i][j] means the number of ways to place first i+1 numbers with the i-th number being j
        dp = [[0] * (n + 1) for _ in range(n + 1)]
        
        # Base case: the first number can be any number from 0 to n
        for j in range(n + 1):
            dp[0][j] = 1
        
        # Fill the dp table
        for i in range(1, n + 1):
            if s[i - 1] == 'I':
                # If s[i-1] is 'I', we want increasing, so accumulate from left to right
                cumulative_sum = 0
                for j in range(n - i + 1):
                    cumulative_sum = (cumulative_sum + dp[i - 1][j]) % MOD
                    dp[i][j] = cumulative_sum
            else:
                # If s[i-1] is 'D', we want decreasing, so accumulate from right to left
                cumulative_sum = 0
                for j in range(n - i, -1, -1):
                    cumulative_sum = (cumulative_sum + dp[i - 1][j + 1]) % MOD
                    dp[i][j] = cumulative_sum
        
        # The answer is the sum of all ways to arrange n+1 numbers
        return dp[n][0]

# Example usage:
# sol = Solution()
# print(sol.numPermsDISequence("DID"))  # Output: 5
17600