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 <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <random>
#include <cmath>
#include <chrono>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cassert>
using namespace std;
using LL = long long;
#define eps 1e-8
#define fi first
#define se second
#define eb emplace_back
#define close std::ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define FOR(i, x, y) for (LL i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (LL i = (x), _##i = (y); i > _##i; --i)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define CASET int ___T; scanf("%d", &___T); for(int __CS=1;__CS<=___T;__CS++)
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
mt19937 dlsrand(random_device{}());
mt19937 mrand(std::chrono::system_clock::now().time_since_epoch().count()); 
int rnd(int x) { return mrand() % x;}
LL bin(LL x, LL n, LL MOD) {LL ret = MOD != 1;for (x %= MOD; n; n >>= 1, x = x * x % MOD)if (n & 1) ret = ret * x % MOD;return ret;}
inline LL get_inv(LL x, LL p) { return bin(x, p - 2, p); }
const double PI = acos(-1.0);
constexpr int maxn = 3050;
constexpr int INF = 0x3f3f3f3f;
constexpr ll linf = 0x3f3f3f3f3f3f3f3f;
constexpr int mod = 1e9+7;
void add(int &a,int b) {a+=b; if(a>=mod) a-=mod;}
int n , dp[maxn][maxn], ans;
char c[maxn];
//dp[i][j]  安排完前i个数第i个数是j的方案
int main()
{
    cin >> n;
    scanf("%s",c+1);
    dp[1][1] = 1;
    for(int i = 2; i <= n; i++){
        if(c[i-1] == '<'){
            for(int j = 2; j <= i; j++){
                dp[i][j] = (dp[i][j-1]+dp[i-1][j-1]) % mod;
            }
        }else {
            for(int j = i-1; j ; j--){
                dp[i][j] = (dp[i][j+1]+dp[i-1][j]) % mod;
            }
        }
    }
    for(int i = 1; i <= n; i++){
        //add(ans, dp[n][i]);
      ans=(ans+dp[n][i])%mod;
    }
    cout << ans << endl;
}