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>
#define DUMP(x)  cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define IFOR(i, m, n) for(int i = n - 1; i >= m; i-- )
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define FOREACH(x,a) for(auto& (x) : (a) )
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
const ll INF = 1LL << 60;
const int MOD = 1000000007;
/* テンプレートここまで */

/* 二項係数 C(n,k) mod p を計算 */
const int MAX = 100010;
long long fac[MAX], finv[MAX], inv[MAX];

void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++){
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}

long long COM(int n, int k){
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main() {
  int N;
  cin >> N;

  string s;
  cin >> s;

  vector<vector<ll>> dp(N+1, vector<ll>(N+1));
  FOR(j, 1, N+1){
    dp[1][j] = 1;
  }

  vector<vector<ll>> sum(N+1, vector<ll>(N+1, 0));
  FOR(j, 1, N+1){
    sum[1][j] = j;
  }

  FOR(i, 2, N+1){
    FOR(j, 1, N-i+2){
      if ( s[i-2] == '<'){
        dp[i][j] = sum[i-1][j] % MOD;
      }else{
        dp[i][j] = (sum[i-1][N-i+2] - sum[i-1][j] +  MOD) %MOD;
        dp[i][j] %= MOD;
      }
      sum[i][j] = sum[i][j-1] + dp[i][j] % MOD;
      sum[i][j] %= MOD; 
    }
  }

  cout << dp[N][1]%MOD << endl;
}