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 (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

const int Mod = 1000000007;

long long div_mod(long long x, long long y, long long z)
{
	if (x % y == 0) return x / y;
	else return (div_mod((1 + x / y) * y - x, (z % y), y) * z + x) / y;
}

char s[3001];
long long dp[3001][3001], fact[3001], fact_inv[3001];

long long recursion(int l, int r)
{
	if (dp[l][r] >= 0) return dp[l][r];
	else dp[l][r] = 0;
	
	int i;
	if (s[l] == '<') dp[l][r] += recursion(l + 1, r);
	if (s[r-1] == '>') dp[l][r] += recursion(l, r - 1);
	for (i = l + 1; i < r; i++) {
		if (s[i-1] == '>' && s[i] == '<') dp[l][r] = (dp[l][r] + recursion(l, i - 1) * recursion(i + 1, r) % Mod * fact[r-l] % Mod * fact_inv[i-l] % Mod * fact_inv[r-i]) % Mod;
	}
	return dp[l][r];
}

int main()
{
	int N;
	scanf("%d", &N);
	scanf("%s", &(s[1]));
	
	int i, j;
	for (i = 1; i <= N; i++) for (j = i + 1, dp[i][i] = 1; j <= N; j++) dp[i][j] = -1;
	for (i = 1, fact[0] = 1; i <= N; i++) fact[i] = fact[i-1] * i % Mod;
	for (i = N - 1, fact_inv[N] = div_mod(1, fact[N], Mod); i >= 0; i--) fact_inv[i] = fact_inv[i+1] * (i + 1) % Mod;
	printf("%lld\n", recursion(1, N));
	fflush(stdout);
	return 0;
}