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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

use std::error::Error;
use std::io;

pub fn main() {
    if let Err(_e) = run() {
        panic!("failed.");
    }
}

fn run() -> Result<(), Box<Error>> {
    let mut input = String::new();
    io::stdin().read_line(&mut input);
    let n = input.trim().parse::<usize>().unwrap();
    input.clear();
    io::stdin().read_line(&mut input);
    let s = input.trim().to_owned();
    input.clear();

    //solve
    let mut dp = vec![vec![0; n]; n];

    for i in 0..n as usize {
        dp[0][i] = 1;
    }

    let mut chars = s.chars();
    for i in 1..n as usize {
        let c = chars.next().unwrap();
        for j in 0..n - i {
            if c == '>' {
                // dp[i][j] = dp[i-1][j+1] + .... dp[i-1][n-i];
                for k in j + 1..n - i + 1 {
                    dp[i][j] += dp[i - 1][k];
                    dp[i][j] %= 1_000_000_007;
                }
            } else {
                // dp[i][j] = dp[i-1][0]...dp[i-1][j], include j
                for k in 0..(j + 1) {
                    dp[i][j] += dp[i - 1][k];
                    dp[i][j] %= 1_000_000_007;
                }
            }
        }
    }
    println!("{}", dp[n - 1][0]);
    Ok(())
}