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>
#define rep(i,N) for(int i=0;i<(int)N;i++)
const int MOD=998244353;
inline int Pow(int x,int y){int z=1;x%=MOD;while(y){if(y&1){z=(1ll*z*x)%MOD;}x=(1ll*x*x)%MOD;y>>=1;}return z;}
int main(void)
{
  int A,B,C,D,ans=0,sum,dp[3000]={0};
  if(!scanf("%d %d %d %d",&A,&B,&C,&D))return 0;
  if(A==C){return printf("%d\n",Pow(A,D-B)),0;}
  else if(B==D){return printf("%d\n",Pow(B,C-A)),0;}
  dp[A-1]=1;
  rep(i,D-B)
  {
    sum=0;
    rep(j,C)
    {
      sum=(1ll*sum*(B+i)%MOD+dp[j])%MOD;
      dp[j]=(1ll*j*dp[j]+sum)%MOD;
    }
  }
  rep(i,C){ans=(ans+1ll*dp[C-i-1]*Pow(D,i))%MOD;}
  return printf("%d\n",ans),0;
}