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)

#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Bytes, Chars, Isize1, Usize1},
};

const MOD: usize = 998244353;

fn main() {
    input! {a:usize,b:usize,c:usize,d:usize}
    let mut dp = vec![vec![1; d + 1]; c + 1];

    for i in a + 1..=c {
        dp[i][b] *= dp[i - 1][b] * b;
        dp[i][b] %= MOD;
    }
    for i in b + 1..=d {
        dp[a][i] = dp[a][i - 1] * a;
        dp[a][i] %= MOD;
    }

    for i in a + 1..=c {
        for j in b + 1..=d {
            dp[i][j] = dp[i - 1][j] * j + dp[i][j - 1] * i;
            dp[i][j] %= MOD;
            dp[i][j] += MOD;
            dp[i][j] -= ((i - 1) * (j - 1) * dp[i - 1][j - 1]) % MOD;
            dp[i][j] %= MOD;
        }
    }
    println!("{}", dp[c][d]);
}