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, fastout};
static MODC: i64 = 998244353;

#[fastout]
fn main() {
    input! {
        a: usize, b: usize, c: usize, d: usize,
    }
    let mut dp: Vec<Vec<i64>> = vec![vec![0; d-b+1]; c-a+1];
    dp[0][0] = 1;
    for v in 0..(c-a+1) {
        for h in 0..(d-b+1) {
            if h != 0 { dp[v][h] = dp[v][h-1]*(a+v)as i64 % MODC }
            if v != 0 { dp[v][h] += dp[v-1][h]*(b+h)as i64 % MODC }
            if h != 0 && v != 0 { dp[v][h] -= dp[v-1][h-1]*((a+v-1)*(b+h-1))as i64 % MODC - MODC }
            dp[v][h] %= MODC
        }
    }
    println!("{}", dp[c-a][d-b]);
}