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 <cstdio>
#include <vector>

const long long mod=998244353;
  
int main()
{
  int a,b,c,d;
  scanf("%d%d%d%d",&a,&b,&c,&d);
  std::vector<std::vector<long long> > dp(2,std::vector<long long>(d-b+1));
  bool F=false;
  for(int i=0;i<=c-a;i++,F=!F){
    for(int j=0;j<=d-b;j++){
      if(i==0){
	if(j==0){
	  dp[F][j]=1ll;
	}
	else{
	  dp[F][j]=dp[F][j-1]*a%mod;
	}
      }
      else{
	dp[F][j]=dp[!F][j]*(b+j)%mod;
	if(j>0){
	  dp[F][j]=(dp[F][j]+(dp[F][j-1]+mod-dp[!F][j-1]*(b+j-1)%mod)*(a+i)+dp[!F][j-1]*(b+j-1))%mod;
	}
      }
      //printf("%d %d %lld\n",i,j,dp[F][j]);
    }
  }
  printf("%lld\n",dp[!F][d-b]);
  return 0;
}