<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:
    string S;
    int mod;
    vector<vector<vector<int> > > dp;
    int rec(int i,int above,int below){
        if(i==S.length()) return 1;
        if(dp[i][above][below]!=-1) return dp[i][above][below];
        long long ans = 0;
        if(S[i]=='I'){
            for(int j = 0;j<above;++j){
                ans = (ans+rec(i+1,above-1-j,below+j))%mod;
            }
        }else{
            for(int j = 0;j<below;++j){
                ans = (ans + rec(i+1,above+j,below-1-j))%mod; 
            }
        }
        return dp[i][above][below] = ans;
    }
    int numPermsDISequence(string s) {
        S = s;
        mod = 1e9+7; 
        int n = s.length();
        dp  = vector<vector<vector<int> > >(n+2,vector<vector<int> > (n+2,vector<int>(n+2,-1)));
        long long ans = 0;
        for(int i = 0;i<=n;++i){
            ans = (ans+rec(0,n-i,i))%mod;
        }
        int temp = ans;
        return temp;
    }
    /*
    0 1 2
    I I - (0,1,2)  
    I D - (1,2,0), (0,2,1)
    D I - (1,0,2), (2,0,1)
    D D - (2,1,0)
    0 1 2 3
    I I I - (0,1,2,3)
    I I D - (0,1,3,2),(0,1,3,2),,(1,2,3,0)
    I D I - (0,2,1,3),(0,3,1,2),,(1,2,0,3),(1,3,0,2)
    
    (i,above,below)
    */
};
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