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: Accepted)

#include<stdio.h>
long long d[3010][3010],M=998244353;
int main(){
  int a,b,c,e,i,j;
  long long x,y,z;
  scanf("%d %d %d %d",&a,&b,&c,&e);
  d[a][b]=1;
  for(i=a;i<c;i++)d[i+1][b]=(d[i][b]*b)%M;
  for(j=b;j<e;j++)d[a][j+1]=(d[a][j]*a)%M;
  for(i=a;i<c;i++){
    for(j=b;j<e;j++){
      x=(d[i+1][j]*(i+1))%M;
      y=(d[i][j+1]*(j+1))%M;
      z=(d[i][j]*i*j)%M;
      d[i+1][j+1]=(x+y+M-z)%M;
    }
  }
  printf("%lld\n",d[c][e]);
  return 0;
}