<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 {
public:
    const int mod=1e9+7;
    int func(int i,int pre,int&n,string&s,unordered_set<int>&s1,vector<vector<int>>&dp){
        if(i==-1){
            if(dp[i+1][pre]!=-1) return dp[i+1][pre];
            int ans=0;
            for(int j=0;j<=n;j++){
                s1.insert(j);
                ans=(ans+func(i+1,j,n,s,s1,dp))%mod;
                s1.erase(j);
            }
            return dp[i+1][pre]=ans;
        }else if(i<n){
            if(dp[i+1][pre]!=-1) return dp[i+1][pre];
            int ans=0;
            if(s[i]=='D'){
                for(int j=0;j<pre;j++){
                    if(s1.find(j)==s1.end()){
                        s1.insert(j);
                        ans=(ans+func(i+1,j,n,s,s1,dp))%mod;
                        s1.erase(j);
                    }
                }
            }else{
                for(int j=pre+1;j<=n;j++){
                    if(s1.find(j)==s1.end()){
                        s1.insert(j);
                        ans=(ans+func(i+1,j,n,s,s1,dp))%mod;
                        s1.erase(j);
                    }
                }
            }
            return dp[i+1][pre]=ans;
        }else{
            return 1;
        }
    }
    int numPermsDISequence(string s) {
        int n=s.length();
        unordered_set<int> s1;
        vector<vector<int>> dp(n+1,vector<int>(n+1,-1));
        return func(-1,0,n,s,s1,dp);
    }
};
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