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:
// M
#include<bits/stdc++.h>
using namespace std;
const int N = 3003, Mod = 998244353;
int n, m, a, b, dp[N][N][2];
int main()
{
	scanf("%d%d%d%d", &a, &b, &n, &m);
	dp[a][b][0] = 1;
	for (int i = a; i <= n; i ++)
		for (int j = b; j <= m; j ++)
			if (i > a || j > b)
			{
				// dp[i][j][0] : i * j and we cannot take the last row
				dp[i][j][0] = (dp[i][j - 1][1] + dp[i][j - 1][0] * 1LL * i) % Mod;
				dp[i][j][1] = (dp[i - 1][j][0] + dp[i - 1][j][1]) * 1LL * j % Mod;
			}
	return !printf("%d\n", (dp[n][m][0] + dp[n][m][1]) % Mod);
}