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;

typedef long long LL;

const int N = 3005;
const int MOD = 998244353;

int A, B, C, D, f[N][N][2];

int main()
{
	scanf("%d%d%d%d", &A, &B, &C, &D);
	f[A][B][1] = 1;
	for (int i = A; i <= C; i++)
		for (int j = B; j <= D; j++) if (i > A || j > B)
		{
			f[i][j][0] = (LL)(f[i - 1][j][0] + f[i - 1][j][1]) * j % MOD;
			f[i][j][1] = (f[i][j - 1][0] + (LL)f[i][j - 1][1] * i % MOD) % MOD;
		}
	printf("%d\n", (f[C][D][0] + f[C][D][1]) % MOD);
	return 0;
}