Problem: 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.

Examples

Input

1 1 2 2


Output

3


Input

2 1 3 4


Output

65


Input

31 41 59 265


Output

387222020

Solution (in CPP):

Solution:
#include<stdio.h>
int f[3001][3001];
const int MOD=998244353;
int main(void)
{
  int A,B,C,D,ans=0,temp;
  if(!scanf("%d %d %d %d",&A,&B,&C,&D))return 0;
  f[A][B]=1;
  for(int i=A+1;i<=C;i++)
  {
    temp=0;
    for(int j=B;j<=D;j++)
    {
      f[i][j]=(temp+1ll*f[i-1][j]*j)%MOD;
      temp=(1ll*(temp+f[i-1][j])*(i-1))%MOD;
    }
  }
  for(int j=B;j<=D;j++){ans=(1ll*ans*C+f[C][j])%MOD;}
  return printf("%d\n",ans),0;
}