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 = 3005;
const long long mod = 998244353ll;

long long f[N][N];
int a, b, c, d;

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] = 1ll;
			else f[i][j] = 0ll;
			if (i > a) f[i][j] = (f[i][j] + f[i - 1][j] * j) % mod;
			if (j > b) f[i][j] = (f[i][j] + f[i][j - 1] * i) % mod;
			if (i > a && j > b) f[i][j] = (f[i][j] + mod * mod - f[i - 1][j - 1] * (i - 1) % mod * (j - 1)) % mod;
		}
	}
	printf("%lld\n", f[c][d]);
	return 0;
}