<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>

constexpr int MAX = 256;
constexpr int MOD = 1000000007;

int dp[MAX];

int Add(int a, int b, int p = MOD) {
  int c = a + b;
  return c < p ? c : c - p;
}

int Sub(int a, int b, int p = MOD) {
  int c = a - b;
  return c < 0 ? c + p : c;
}

class Solution {
public:
    int numPermsDISequence(string s) {
      int n = s.length();
      fill(dp, dp + n + 1, 0);
      dp[0] = 1;
      for (int i = 1; i <= n; ++i) {
        bool d = s[i - 1] == 'D';
        int sum = dp[i - 1];
        for (int j = i; j > 0; --j) {
          dp[j] = d ? Sub(sum, dp[j - 1]) : dp[j - 1];
        }
        dp[0] = d ? sum : 0;
        for (int j = 1; j <= i; ++j) {
          dp[j] = Add(dp[j], dp[j - 1]);
        }
      }
      return dp[n];
    }
};
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