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 maxn = 3010;
const int mod = 998244353;

int s, t, n, m;
int f[maxn][maxn];

int main() {
	scanf("%d%d%d%d", &s, &t, &n, &m);
	f[s][t] = 1;
	for (int i = s; i <= n; i++) {
		for (int j = t; j <= m; j++) {
			if (i == s && j == t) continue;
			int v1 = 1LL * j * f[i-1][j] % mod;
			int v2 = 1LL * i * f[i][j-1] % mod;
			int v3 = 1LL * (i-1) * (j-1) % mod * f[i-1][j-1] % mod;
			f[i][j] = ((v1 + v2) % mod + mod - v3) % mod;
		}
	}
	printf("%d\n", f[n][m]);
	return 0;
}