<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>

#define ll long long
class Solution {
public:
    vector<bool> mark;
    int mod = 1e9 + 7;

    int solve(string& s, int index, int prev, vector<vector<int>>& dp) {
        int n = s.length();
        if (index == n) {
            return 1;
        }
        
        if (dp[index][prev] != -1)
            return dp[index][prev];
        int ans = 0;
        int temp = (s[index] == 'D' ? -1 : 1);
        int i = prev + temp;
        while (i <= n && i >= 0) {
            if (!mark[i]) {
                mark[i] = 1;
                ans = (ans + solve(s, index + 1, i, dp)) % mod;
                mark[i] = 0;
            }
            i += temp;
        }
        return dp[index][prev] = ans;
    }
    int numPermsDISequence(string s) {
        int n = s.length();
        mark.resize(s.length() + 1, false);
        vector<vector<int>> dp(n + 1, vector<int>(n + 1, -1));
        int ans = 0;
        for (int i = 0; i <= n; i++) {
            mark[i] = true;
            ans = (ans + solve(s, 0, i, dp)) % mod;
            mark[i] = false;
        }
        return ans;
    }
};
10500