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 <stdio.h>
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 3005;
const ll MOD = 998244353;

int A,B,C,D;

ll f[MAXN][MAXN];

int main()
{
	cin >> A >> B >> C >> D;
	for (int i = A;i <= C;i++)
		for (int j = B;j <= D;j++)
		{
			if (i == A && j == B)
			{
				f[i][j] = 1;
				continue;
			}
			if (i == A)
				f[i][j] = f[i][j - 1] * i % MOD;
			if (j == B)
				f[i][j] = f[i - 1][j] * j % MOD;
			if (i != A && j != B)
				f[i][j] = (f[i][j - 1] * i + f[i - 1][j] * j - f[i - 1][j - 1] * (i - 1) % MOD * (j - 1)) % MOD;
		}
	cout << (f[C][D] + MOD) % MOD << endl;
	return 0;
}