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>
using namespace std;
const int N=3011;
const int MOD=1e9+7;

int n;
char s[N];
int dp[N][N],sum[N];
void Add(int &x,int y)
{
	x+=y;
	if(x>=MOD) x-=MOD;
}

int main()
{
	
	scanf("%d%s",&n,&s);
	for(int i=1;i<=n;i++)
		dp[i-1][n-i]=1;
	for(int len=n-1;len>0;len--)
	{
		bool ok=(s[n-len-1]=='<');
		if(ok) for(int i=0;i<=len;i++) Add(dp[i][len-i-1],dp[i][len-i]);
		if(!ok) for(int i=1;i<=len;i++) Add(dp[i-1][len-i],dp[i][len-i]);
		if(ok) for(int i=1;i<=len;i++) Add(dp[i][len-i-1],dp[i-1][len-i]);
		if(!ok) for(int i=len-1;i>=0;i--) Add(dp[i][len-i-1],dp[i+1][len-i-2]);
	}
	printf("%d\n",dp[0][0]);
	
	return 0;
}