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>
#define SIZE 3005
#define MOD 998244353
using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

ll dp[SIZE][SIZE];
int main()
{
	int A,B,C,D;
	scanf("%d %d %d %d",&A,&B,&C,&D);
	for(int i=A;i<=C;i++)
	{
		for(int j=B;j<=D;j++)
		{
			if(i==A&&j==B) dp[i][j]=1; 
			else if(i==A) dp[i][j]=dp[i][j-1]*(ll) A%MOD;
			else if(j==B) dp[i][j]=dp[i-1][j]*(ll) B%MOD;
			else
			{
				dp[i][j]+=dp[i-1][j]*(ll) j;
				dp[i][j]+=dp[i][j-1]*(ll) i;
				dp[i][j]%=MOD;
				dp[i][j]-=dp[i-1][j-1]*(ll) (i-1)*(ll) (j-1)%MOD;
				if(dp[i][j]<0) dp[i][j]+=MOD;
			}
		}
	}
	printf("%lld\n",dp[C][D]);
}