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 rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair

using namespace std;

const LL MOD=998244353;

LL a,b,c,d,dp[3010][3010];

int main()
{
	cin>>a>>b>>c>>d;
	dp[a][b]=1;
	for(int i=a+1;i<=c;i++) dp[i][b]=dp[i-1][b]*b%MOD;
	for(int i=b+1;i<=d;i++) dp[a][i]=dp[a][i-1]*a%MOD;
	for(int i=a+1;i<=c;i++) for(int j=b+1;j<=d;j++)
	dp[i][j]=(dp[i-1][j]*j+dp[i][j-1]*i-dp[i-1][j-1]*(i-1)*(j-1)%MOD+MOD)%MOD;
	cout<<dp[c][d]<<endl;
	return 0;
}