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;
//mt19937 mrand(chrono::high_resolution_clock::now().time_since_epoch().count());
//int rnd(int x) { return mrand()%x;}
typedef long long ll;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll dp[3100][3100];
const ll md=998244353;
int main(){
	dp[0][0]=1;
	ll a,b,c,d;
	scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
	for(int i=1;i<3100;i++){
		dp[0][i]=dp[0][i-1]*a%md;
		dp[i][0]=dp[i-1][0]*b%md;
	}
	for(int i=1;i<=c-a;i++){
		for(int j=1;j<=d-b;j++){
			dp[i][j]=dp[i-1][j]*(j+b)%md+dp[i][j-1]*(i+a)%md-(dp[i-1][j-1]*(i-1+a)%md*(j-1+b)%md)%md+md;
			dp[i][j]%=md;
		}
	}
	printf("%lld\n",dp[c-a][d-b]);
	return 0;
}