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;
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long

const int N = 3005;
const int MOD = 998244353;

int a, b, c, d;
int cache[N][N];

int dp(int x, int y)
{
	if(x > c)
		return 0;
	if(y > d)
		return 0;
	if(x == c && y == d)
		return 1;
	int &ans = cache[x][y];
	if(ans != -1)
		return ans;
	ans = 0;
	ans += y * dp(x + 1, y) + x * dp(x, y + 1) - x * y * dp(x + 1, y + 1);
	ans %= MOD;
	ans += MOD;
	ans %= MOD;
	return ans;
}

int32_t main()
{
	IOS;
	memset(cache, -1, sizeof(cache));
	cin >> a >> b >> c >> d;
	int ans = dp(a, b);
	cout << ans;
	return 0;
}