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 mod = 998244353;
int A, B, C, D;
int f[3005][3005];
int main() {
	scanf("%d%d%d%d", &A, &B, &C, &D);
	f[A][B] = 1;
	for (int i = A + 1; i <= C; i++) {
		int sc = 0;
		for (int j = B; j <= D; j++) {
			sc = 1ll * sc * (i - 1) % mod;
			f[i][j] = (1ll * f[i - 1][j] * j + sc) % mod;
			sc = (sc + f[i - 1][j]) % mod;
		}
	}
	int sc = 0;
	for (int j = B; j <= D; j++) {
		sc = (1ll * sc * C + f[C][j]) % mod;
	}
	printf("%d\n", sc);
	return 0;
}