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 endl '\n'
#define pb push_back
#define mp make_pair
#define int long long
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
const int MOD=998244353;
int a,b,c,d;
int dp[3010][3010];//
signed main(){
	memset(dp,0,sizeof dp);
	cin>>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){
			dp[i][j]=(dp[i-1][j]*j+dp[i][j-1]*i+MOD-dp[i-1][j-1]*(i-1)*(j-1)+MOD)%MOD;
//			cerr<<i<<' '<<j<<' '<<dp[i][j]<<endl;
			if(dp[i][j]<0)dp[i][j]+=MOD;
		}
	cout<<dp[c][d]<<endl;
	return 0;
}