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: Runtime Error)

#include <iostream>
#include <cmath>

using namespace std;


const int mod = 998244353;
const int maxSize = 3000;
int sum[maxSize][maxSize];

int main(){
    //grid AxB
    int A,B,C,D;
    cin >> A >> B >> C >> D;
    int a,b;
    sum[A][B]=0;
    sum[A][B+1]=1;
    sum[A+1][B]=1;
    for(a = A+1; a <= C; a++){
        for(b = B+1; b<=D; b++){
            sum[a][b]= ((sum[a][b-1]*a)+(sum[a-1][b]*b) -1 ) %mod;
        }
    }

    cout << sum[C][D];





}