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 rep(i,a,b) for(int i=(a);i<=(b);++i)
#define rrep(i,a,b) for(int i=(a);i>=(b);--i)
#define PB push_back
#define ar2 array<int, 2>
typedef long long LL;
const LL P = 998244353;
const int N = 3005;
mt19937 rng(time(0));

LL dp[N][N];
int a,b,c,d;

int main(){
	cin>>a>>b>>c>>d;
	
	dp[a][b] = 1;
	rep(i,a,c) rep(j,b,d){
		if(i==a && j==b) continue;
		if(i==a) dp[i][j] = dp[i][j-1]*i%P;
		else if(j==b) dp[i][j] = dp[i-1][j]*j%P;
		else{
			dp[i][j] = (dp[i][j-1]*i%P + dp[i-1][j]*j%P - dp[i-1][j-1]*(i-1)%P*(j-1)%P + P) % P;
		}
	}
	cout<<dp[c][d]<<endl;
	return 0;
}