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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main() {
  int N; cin >> N;
  string s; cin >> s;
  const int MOD = 1000000007;
  // dp[i][j] := 左から i 番目まで決めたとき、
  // 残り N-i 個のうち j 個が左から i 番目の数字より小さいような場合の数
  vector<vector<int>> dp(N+1, vector<int>(N+1, 0));
  dp.at(0).at(0) = 1;
  for (int i = 0; i < N; i++) dp.at(1).at(i) = 1;
  for (int i = 1; i < N; i++) {
    vector<long long> S(N+2);
    // 累積和
    for (int j = 0; j < N+1; j++) {
      S.at(j+1) = (S.at(j) + dp.at(i).at(j)) % MOD;
    }
    if (s.at(i-1) == '<') {
      for (int j = 0; j <= N-i-1; j++) {
        dp.at(i+1).at(j) += S.at(j+1);
        dp.at(i+1).at(j) % MOD;
      }
    }
    else {  // s[i] == '>'
      for (int j = 0; j < N; j++) {
        dp.at(i+1).at(j) += S.at(N-i+1) - S.at(j+1);
        dp.at(i+1).at(j) % MOD;
      }
    }
  }
  cout << dp.at(N).at(0) << endl;
}