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;
	using ll = long long;
	#define forn(i,n) for(int i = 0; i< n;++i)
	int INF = 1e9 + 7;
	ll MOD = 998244353;


	int main(){
		ios_base::sync_with_stdio(false);
		cin.tie(NULL);
		int a,b,c,d;
		cin >> a >> b >> c >> d;

		vector<vector<ll>> dp(c+10,vector<ll>(d+10,0));
		
		for(int i = a; i <= c; ++i){
			for(int j = b; j <= d; ++j){
				if(i == a && j == b){
					dp[i][j] = 1;
					continue;
				}
				dp[i][j] = dp[i-1][j]*j + dp[i][j-1]*i - dp[i-1][j-1]*(i-1)*(j-1);
				//dp[i][j] %= MOD;
				dp[i][j] = (MOD + dp[i][j] % MOD) % MOD;
			}
		}
		cout << dp[c][d] << endl;
				
	}