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)

use std::io::*;
const MOD: usize = 998244353;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let a: usize = itr.next().unwrap().parse().unwrap();
    let b: usize = itr.next().unwrap().parse().unwrap();
    let c: usize = itr.next().unwrap().parse().unwrap();
    let d: usize = itr.next().unwrap().parse().unwrap();
    let n = c - a;
    let m = d - b;

    let mut dp = vec![vec![0; m + 1]; n + 1];
    dp[0][0] = 1;
    for i in 0..=n {
        for j in 0..=m {
            if i > 0 {
                dp[i][j] += (b + j) * dp[i - 1][j] % MOD;
            }
            if j > 0 {
                dp[i][j] += (a + i) * dp[i][j - 1] % MOD;
            }
            if i > 0 && j > 0 {
                dp[i][j] += MOD;
                dp[i][j] -= (a + i - 1) * (b + j - 1) * dp[i - 1][j - 1] % MOD;
            }
            dp[i][j] %= MOD;
        }
    }

    println!("{}", dp[n][m]);
}