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>
using namespace std;
int A, B, C, D;
long long dptable[3001][3001];
const long long mod = 998244353;
long long dp(int i, int j) {
	if (i == A && j == B) return 1;
	if (dptable[i][j] != 0) return dptable[i][j];
	if (i == A) return dptable[i][j] = i * dp(i, j - 1) % mod;
	if (j == B) return dptable[i][j] = j * dp(i - 1, j) % mod;
	return dptable[i][j] = (i * dp(i, j - 1) + j * dp(i - 1, j) - (i - 1) * (j - 1) * dp(i - 1, j - 1)) % mod;
}
int main() {
	scanf("%d %d %d %d", &A, &B, &C, &D);
	printf("%lld", (dp(C, D) + mod) % mod);
	return 0;
}