Problem:
Score : 600 points

Problem Statement

We have a grid with A horizontal rows and B vertical columns, with the squares painted white. On this grid, we will repeatedly apply the following operation:

Assume that the grid currently has a horizontal rows and b vertical columns. Choose "vertical" or "horizontal".

If we choose "vertical", insert one row at the top of the grid, resulting in an (a+1) \times b grid.

If we choose "horizontal", insert one column at the right end of the grid, resulting in an a \times (b+1) grid.

Then, paint one of the added squares black, and the other squares white.

Assume the grid eventually has C horizontal rows and D vertical columns. Find the number of ways in which the squares can be painted in the end, modulo 998244353.

Constraints

1 \leq A \leq C \leq 3000

1 \leq B \leq D \leq 3000

A, B, C, and D are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the number of ways in which the squares can be painted in the end, modulo 998244353.

Sample Input 1

1 1 2 2

Sample Output 1

3

Any two of the three squares other than the bottom-left square can be painted black.

Sample Input 2

2 1 3 4

Sample Output 2

65

Sample Input 3

31 41 59 265

Sample Output 3

387222020

Solution in Rust (1.42.0) (judge verdict: Accepted)

fn main() {
    const MOD: usize = 998244353;

    let [a, b, c, d]: [usize; 4] = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let mut iter = buf.split_whitespace();
        [
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
        ]
    };

    let mut dp = vec![vec![[0; 2]; d + 1]; c + 1];
    dp[a][b][1] = 1;
    for i in a..=c {
        for j in b..=d {
            if i + 1 <= c {
                dp[i + 1][j][0] = (dp[i][j][0] + dp[i][j][1]) * j % MOD;
            }
            if j + 1 <= d {
                dp[i][j + 1][1] = (dp[i][j][0] + dp[i][j][1] * i) % MOD;
            }
        }
    }

    let ans = (dp[c][d][0] + dp[c][d][1]) % MOD;
    println!("{}", ans);
}