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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int A,B,C,D;
const int Mod=998244353;
#define Maxn 3005
int dp[Maxn][Maxn][2];

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