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 std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use num_traits::clamp;
use ordered_float::OrderedFloat;
use proconio::{input, marker::*, fastout};
use superslice::*;

const M: usize = 998244353;

#[fastout()]
fn main() {
    input! {
        a: usize, b: usize, c: usize, d: usize,
    }

    let mut dp = vec![vec![vec![0usize; d+1]; c+1]; 2];
    dp[1][a][b] = 1;
    for y in a..=c {
        for x in b..=d {
            if y != a {
                dp[0][y][x] = (dp[0][y-1][x] + dp[1][y-1][x]) * x;
                dp[0][y][x] %= M;
            }
            if x != b {
                dp[1][y][x] = dp[0][y][x-1] + dp[1][y][x-1] * y;
                dp[1][y][x] %= M;
            }
        }
    }
    let ans = (dp[0][c][d] + dp[1][c][d]) % M;
    println!("{}", ans);
}