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;
#define ll long long
const int mod=998244353;
int read() {
	int x=0, f=1; char ch=' ';
	while(!isdigit(ch)) {ch=getchar(); if(ch=='-') f=-1;}
	while(isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48), ch=getchar();
	return x*f;
}
ll f[3005][3005];
int main() {
	int a, b, c, d;
	cin>>a>>b>>c>>d;
	f[a][b]=1;
	for(int i=a;i<=c;i++) {
		for(int j=b;j<=d;j++) {
			if(i>a) f[i][j]+=f[i-1][j]*j;
			if(j>b) f[i][j]+=f[i][j-1]*i;
			if(i>a&&j>b) f[i][j]-=(ll)(i-1)*(j-1)*f[i-1][j-1];
			f[i][j]%=mod;
		}
	}
	cout<<(f[c][d]%mod+mod)%mod;
}