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)

#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

const int MOD = 1000 * 1000 * 1000 + 7;

void add(int& x, int y) {
    x += y;
    if (x >= MOD) {
        x -= MOD;
    }
}

const int MX = 3007;
int f[MX][MX], sf[MX][MX];

int main() {
#ifdef FASTIO
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
    int n;
    cin >> n;
    string s;
    cin >> s;
    s = "$$" + s;
    f[1][1] = 1;
    for (int i = 1; i < MX; i++) {
        sf[1][i] = 1;
    }
    for (int i = 2; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            if (s[i] == '<') {
                f[i][j] = sf[i - 1][j - 1];
            } else {
                f[i][j] = (sf[i - 1][MX - 1] - sf[i - 1][j - 1] + MOD) % MOD;
            }
        }
        for (int j = 1; j < MX; j++) {
            sf[i][j] = (sf[i][j - 1] + f[i][j]) % MOD;
        }
    }

    cout << sf[n][MX - 1] << "\n";
    return 0;
}