Problem:
Score : 100 points

Problem Statement

Let N be a positive integer.
You are given a string s of length N - 1, consisting of < and >.

Find the number of permutations (p_1, p_2, \ldots, p_N) of (1, 2, \ldots, N) that satisfy the following condition, modulo 10^9 + 7:

For each i (1 \leq i \leq N - 1), p_i < p_{i + 1} if the i-th character in s is <, and p_i > p_{i + 1} if the i-th character in s is >.

Constraints

N is an integer.

2 \leq N \leq 3000

s is a string of length N - 1.

s consists of < and >.

Input

Input is given from Standard Input in the following format:

N
s

Output

Print the number of permutations that satisfy the condition, modulo 10^9 + 7.

Sample Input 1

4
<><

Sample Output 1

5

There are five permutations that satisfy the condition, as follows:

(1, 3, 2, 4)

(1, 4, 2, 3)

(2, 3, 1, 4)

(2, 4, 1, 3)

(3, 4, 1, 2)

Sample Input 2

5
<<<<

Sample Output 2

1

There is one permutation that satisfies the condition, as follows:

(1, 2, 3, 4, 5)

Sample Input 3

20
>>>><>>><>><>>><<>>

Sample Output 3

217136290

Be sure to print the number modulo 10^9 + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>

using namespace std;

using int64 = long long;
const int INF = 1 << 30;
const int mod = 1e9 + 7;


int main() {
  int N;
  string S;
  cin >> N;
  cin >> S;
  vector< int > dp(N + 1); // dp[i] : 基準点より小さいものの個数がi個あるときのうく
  for(int i = N - 1; i >= 0; i--) dp[i] = 1;

  for(int i = N - 1; i > 0; i--) {
    vector< int > dp2(i + 5);
    for(int j = 0; j <= i; j++) {
      int up_count = j;
      int down_count = i - j;

      if(S[i - 1] == '<') {
        (dp2[0] += dp[j]) %= mod;
        (dp2[up_count] += mod - dp[j]) %= mod;
        /*
        for(int k = 1; k <= up_count; k++) {
          (dp2[up_count - k] += dp[j]) %= mod;
        }
         */
      } else {
        (dp2[up_count] += dp[j]) %= mod;
        (dp2[up_count + down_count] += mod - dp[j]) %= mod;
        /*
        for(int k = 1; k <= down_count; k++) {
          (dp2[up_count + (k - 1)] += dp[j]) %= mod;
        }
         */
      }
    }
    for(int j = 1; j < dp2.size(); j++) {
      (dp2[j] += dp2[j - 1]) %= mod;
    }
    dp.swap(dp2);
  }

  cout << accumulate(begin(dp), end(dp), 0LL) % mod << endl;
}