<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 
{
    const long long int MOD = 1e9 + 7;
    long long int dp[205][205];
    long long int psum[205];
    public:
        int numPermsDISequence(string s) 
        {
            dp[1][1] = 1;
            psum[1] = 1;
            int n = s.length() + 1;
            for (int i = 2; i <= n; i++)
            {
                for (int j = 1; j <= i; j++)
                {
                    if (s[i-2] == 'I')
                    {
                        dp[i][j] += psum[j - 1] % MOD;
                        dp[i][j] %= MOD;
                    }
                    else
                    {
                        long long int val = psum[i - 1] - psum[j - 1];
                        val %= MOD;
                        if (val < 0) val += MOD;
                        dp[i][j] += val;
                        dp[i][j] %= MOD;
                    }
                }
                for (int j = 1; j <= i; j++)
                {
                    psum[j] = psum[j - 1] + dp[i][j];
                    psum[j] %= MOD;
                }
            }

            long long int res = 0;

            for (int i = 0; i <= n; i++)
            {
                res += dp[n][i];
                res %= MOD;
            }

            return res;
        }
};
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