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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn=3050;
const int maxm=3050;
const ll MOD=998244353;
ll dp[maxn][maxm];
int a,b,c,d;

void solve(){
	scanf("%d %d %d %d",&a,&b,&c,&d);
	for(int i=a;i<=c;i++){
		for(int j=b;j<=d;j++){
			if(i==a&&j==b)	dp[i][j]=1;
			else if(i==a)	dp[i][j]=dp[i][j-1]*i%MOD;
			else if(j==b)	dp[i][j]=dp[i-1][j]*j%MOD;
			else{
				dp[i][j]=(dp[i][j-1]*i%MOD+dp[i-1][j]*j%MOD-dp[i-1][j-1]*(i-1)%MOD*(j-1)%MOD+MOD)%MOD;
			}
		}
	}
	printf("%lld\n",dp[c][d]);
}
int main(){
	solve();
	return 0;
}