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 ll long long
using namespace std;
const ll mod = 998244353;
const int maxn = 3100;
ll a, b, c, d;
ll dp[maxn][maxn];

int main() {
	cin>>a>>b>>c>>d;

	dp[a][b] = 1LL;

	for(int i=1;i<=c;i++) {
		for(int j=1;j<=d;j++) {
			if(i == a && j == b) continue;

			dp[i][j] = (i * dp[i][j-1])%mod + (j * dp[i-1][j])%mod; dp[i][j] %= mod;
			dp[i][j] = dp[i][j] + mod - (dp[i-1][j-1] * (i - 1) * (j - 1))%mod; dp[i][j] %= mod;
		}
	}

	cout<<dp[c][d]<<"\n";
}