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 proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::f64::consts::*;

#[allow(unused)]
const INF: usize = std::usize::MAX / 4;
#[allow(unused)]
const M: usize = 998244353;

fn main() {
    input! {
        a: usize,
        b: usize,
        c: usize,
        d: usize,
    }
    let mut dp = vec![vec![0; d + 1]; c + 1];
    dp[a][b] = 1;
    for i in a + 1..=c {
        dp[i][b] = b * dp[i - 1][b] % M;
    }
    for j in b + 1..=d {
        dp[a][j] = a * dp[a][j - 1] % M;
    }
    for i in a + 1..=c {
        for j in b + 1..=d {
            dp[i][j] = ((dp[i - 1][j] * j % M) + (dp[i][j - 1] * i % M) + M
                - ((i - 1) * (j - 1) * dp[i - 1][j - 1]) % M)
                % M;
        }
    }
    println!("{}", dp[c][d]);
}