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>

const int MOD = 998244353, MAXN = 3e3 + 10;
inline int add(int x, int y) { return x + y >= MOD ? x + y - MOD : x + y; }
inline int sub(int x, int y) { return x < y ? x - y + MOD : x - y; }
inline int mul(int x, int y) { return 1LL * x * y - 1LL * x * y / MOD * MOD; }

int A, B, C, D;
int dp[MAXN][MAXN];

int main() {
	scanf("%d%d%d%d", &A, &B, &C, &D);
	dp[A][B] = 1;
	for (int i = A; i <= C; ++i)
		for (int j = B; j <= D; ++j) if (i > A || j > B) dp[i][j] = sub(add(mul(dp[i - 1][j], j), mul(dp[i][j - 1], i)), mul(dp[i - 1][j - 1], mul(i - 1, j - 1)));
	printf("%d\n", dp[C][D]);
	return 0;
}