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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define MOD 1000000007

char s[3010];
ll dp[3010][3010];//dp[i][j]=(i番目まで決めて、i番目より大きいものがj個残っている)

int main(){
	ll n;
	scanf("%lld %s",&n,s);
	rep(j,0,n)dp[1][j]=1;
	rep(i,1,n){
		ll temp[3010]={};
		rep(j,0,n-i+1)temp[j+1]=(temp[j]+dp[i][j])%MOD;
		rep(j,0,n-i){
			if(s[i-1]=='<'){
				//大きいものから使うので、jは減る
				//配るDP
				//rep(k,0,j)dp[i+1][k]=(dp[i+1][k]+dp[i][j])%MOD;
				//もらうDPにして累積和で加速
				dp[i+1][j]=(temp[n-i+1]-temp[j+1])%MOD;
			}else{
				//小さいものから使うので、jは増える
				//配るDP
				//rep(k,j,n-i+1)dp[i+1][k]=(dp[i+1][k]+dp[i][j])%MOD;
				//もらうDP
				dp[i+1][j]=temp[j+1];
			}
		}
	}
	printf("%lld\n",(dp[n][0]+MOD)%MOD);
	return 0;
}