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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main{
	public static void main(String args[]){
    	Scanner sc = new Scanner(System.in);
      
      	int a = sc.nextInt();
      	int b = sc.nextInt();
      	int c = sc.nextInt();
      	int d = sc.nextInt();
      
      	int r_sum = 0;
      	int r = 0;
      
      	int a_a = 1;
        int b_b = 1;
      
      	r = a * (d-b);
      	for(int i = a+1; i <= c; i++){
              r = i - a;
              for(int k = 1; k <= d; k++){
                int j = k;
                for(; j <= d; j++){
                     r *= i; 
                }
                for(j = d - k; j <= d; j++){
                     r *= i - 1; 
                }
              }
          
          	r_sum += r;
	        r = 0;
        }
      System.out.print(r_sum);
      	
    }
}