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;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
#define rep(i, n) for(UL i = 0; i < (n); i++)

const ULL M = 998244353;

ULL dp[3001][3001] = {};

int main() {
	UL w, h, W, H; cin >> w >> h >> W >> H;
	dp[w][h] = 1;
	for (UL x = w; x <= W; x++)
		for (UL y = h; y <= H; y++) {
			if (x == w && y == h) continue;
			ULL tmp = 0;
			tmp += dp[x - 1][y] * y;
			tmp += dp[x][y - 1] * x;
			tmp += M * ((x - 1) * (y - 1));
			tmp -= dp[x - 1][y - 1] * ((x - 1) * (y - 1));
			tmp %= M;
			dp[x][y] = tmp;
		}
	cout << dp[W][H] << endl;
	return 0;
}