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;

const MOD: i64 = 998244353;

fn main() {
    input! {
	A: usize,
	B: usize,
	C: usize,
	D: usize,
    }

    let mut dp = vec![0i64; 3001];
    let mut prevDp = vec![0i64; 3001];

    for i in A..C+1 {
	std::mem::swap(&mut dp, &mut prevDp);
	for j in B..D+1 {
	    dp[j] = 0;
	    if i == A && j == B {
		dp[j] = 1;
		continue;
	    }
	    if i > A {
		dp[j] = (dp[j] + prevDp[j] * j as i64) % MOD;
	    }
	    if j > B {
		dp[j] = (dp[j] + dp[j - 1] * i as i64) % MOD;
	    }
	    if i > A && j > B {
		dp[j] = (dp[j] - prevDp[j - 1] * (i - 1) as i64 * (j - 1) as i64) % MOD
	    }
	}
    }

    println!("{}", (dp[D] + MOD) % MOD);
}