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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

long long dp[3005][3005];
const long long mod = 998244353;

int main()
{
	int a, b, c, d;
	cin >> a >> b >> c >> d;
	dp[a][b] = 1;
	for(int i = a; i <= c; i++){
		for(int j = b; j <= d; j++){
			if(a < i) dp[i][j] = (dp[i][j] + dp[i - 1][j] * j) % mod;
			if(b < j) dp[i][j] = (dp[i][j] + dp[i][j - 1] * i) % mod;
			if(a < i && b < j) dp[i][j] = (dp[i][j] - dp[i - 1][j - 1] * (i - 1) * (j - 1)) % mod;
		}
	}
	cout << dp[c][d] << endl;
	return 0;
}