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<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N=3000,mod=998244353;

int a,b,c,d;
LL dp[N+9][N+9];

void work(){
  scanf("%d%d%d%d",&a,&b,&c,&d);
  dp[a][b]=1;
  for (int i=a;i<=c;++i)
	for (int j=b;j<=d;++j){
	  if (i==a&&j==b) continue;
	  if (i>a) dp[i][j]+=dp[i-1][j]*j;
	  if (j>b) dp[i][j]+=dp[i][j-1]*i;
	  if (i>a&&j>b) dp[i][j]-=dp[i-1][j-1]*(i-1)*(j-1);
	  dp[i][j]=(dp[i][j]%mod+mod)%mod;
	}
  printf("%lld\n",dp[c][d]);
}

int main(){
  int T=1;
  //scanf("%d",&T);
  for (;T--;) work();
  return 0;
}