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 = 3005, mod = 998244353;

int A, B, C, D;
int f[maxN + 1][maxN + 1];

inline void update(int &x, int y) { x = x + y >= mod ? x + y - mod : x + y; }

int main()
{
	scanf("%d %d %d %d", &A, &B, &C, &D);

	f[A][B] = 1;
	for(int i = A; i <= C; i++)
		for(int j = B; j <= D; j++)
		{
			if(i == A && j == B) continue;
			update(f[i][j], 1ll * f[i - 1][j] * j % mod);
			update(f[i][j], 1ll * f[i][j - 1] * i % mod);
			update(f[i][j], mod - 1ll * f[i - 1][j - 1] * (i - 1) * (j - 1) % mod);
		}

	printf("%d\n", f[C][D]);
	return 0;
}