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::{fastout, input};

use std::collections::HashMap;

fn pow(mut x: i64, mut n: i64, p: i64) -> i64 {
    let mut r = 1;
    while n > 0 {
        if n & 1 == 1 {
            r = r * x % p;
        }
        x = x * x % p;
        n >>= 1;
    }
    r
}

#[fastout]
fn main() {
    const M: i64 = 998244353;
    input! {
        a: i64, b: i64, c: i64, d: i64,
    }
    let mut h = vec![vec![None; d as usize + 1]; c as usize + 1];
    fn f(c: i64, d: i64, a: i64, b: i64, h: &mut Vec<Vec<Option<i64>>>) -> i64 {
        if let Some(v) = h[c as usize][d as usize] {
            return v;
        }
        let v = if c == a && d == b {
            1
        } else if c == a {
            f(c, d - 1, a, b, h) * c
        } else if d == b {
            f(c - 1, d, a, b, h) * d
        } else {
            f(c, d - 1, a, b, h) * c + f(c - 1, d, a, b, h) * d
                - f(c - 1, d - 1, a, b, h) * ((c - 1) * (d - 1))
        } % M;
        let v = (v + M) % M;
        h[c as usize][d as usize] = Some(v);
        v
    }
    let ans = f(c, d, a, b, &mut h);
    println!("{}", ans);
}