<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 MOD 1000000007

class Solution {
public:
    vector<vector<vector<long long>>> dp;
    long long n;
    string s;

    long long find(long long loc, long long abv, long long blw){
        if(loc == n+1) return 1;
        if(dp[loc][abv][blw] != -1) return dp[loc][abv][blw];

        long long ans = 0;

        if(s[loc-1] == 'D'){
            for(int i = 0; i < blw; i++){
                ans = ans + find(loc+1, abv + blw -i-1, i);
                ans%=MOD;
            }
        }
        else{
            for(int i = 0; i < abv; i++){
                ans = ans + find(loc+1, abv-i-1, blw + i);
                ans%=MOD;
            }
        }

        return dp[loc][abv][blw] = ans;
    }

    int numPermsDISequence(string S) {
        s = S;
        n = s.length();
        dp = vector<vector<vector<long long>>> (n+2, vector<vector<long long>> (n+3, vector<long long>(n+3, -1)));

        long long fin = 0;

        for(int i = 0; i <= n; i++){
            fin = fin + find(1, n-i, i);
            fin%=MOD;
        }

        return fin;
    }
};
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