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>
#include<stdlib.h>

#define POS(i,j) (((i)&1)*n+(j))

const int mod=1000000007;

void run(void){
  int n;
  scanf("%d",&n);
  char *s=(char *)calloc(n,sizeof(char));
  scanf("%s",s);
  int *dp=(int *)calloc(2*n,sizeof(int));
  int i,j;
  for(i=0;i<n;i++) dp[POS(0,i)]=1;
  for(i=1;i<n;i++){
    for(j=0;j<n;j++) dp[POS(i,j)]=0;
    if(s[i-1]=='<'){
      for(j=0;j<n-i;j++){
	dp[POS(i,j)]+=dp[POS(i-1,j)];
	dp[POS(i,j)]%=mod;
      }
    } else {
      for(j=1;j<n+1-i;j++){
	dp[POS(i,0)]+=dp[POS(i-1,j)];
	dp[POS(i,0)]%=mod;
	dp[POS(i,j)]+=mod-dp[POS(i-1,j)];
	dp[POS(i,j)]%=mod;
      }
    }
    for(j=1;j<n-i;j++){
      dp[POS(i,j)]+=dp[POS(i,j-1)];
      dp[POS(i,j)]%=mod;
    }
  }
  printf("%d\n",dp[POS(n-1,0)]);
}

int main(void){
  run();
  return 0;
}