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;

const int N = 3000 + 10;
const int MOD = 998244353;

int f[N][N];

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