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 <cstdio>
#include <algorithm>

typedef long long LL;
const int Mod = 998244353;
const int MN = 3005;

int A, B, C, D;
int f[MN][MN], g[MN][MN];

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