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<cstdio>
#include<cstring>
using namespace std;
const int mod=998244353;
#define N 3030
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int dp[N][N],A,B,C,D;
int main(){
	A=read(),B=read(),C=read(),D=read();
	dp[A][B]=1;
	for(int i=A;i<=C;++i){
		for(int j=B;j<=D;++j){
			if(i==A&&j==B)continue;
			dp[i][j]=(1LL*dp[i-1][j]*j+1LL*dp[i][j-1]*i-1LL*dp[i-1][j-1]*(i-1)%mod*(j-1)%mod+mod)%mod;
		}
	}
	printf("%d\n",dp[C][D]);
	return 0;
}