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 <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using std::cin;using std::cout;using std::endl;using std::cerr;

typedef long long ll;

const ll mod = 998244353,mmax = 3000;

ll a,b,c,d;
ll dp[mmax + 10][mmax + 10];

int main(){

	std::ios::sync_with_stdio(0);
	cin.tie(0);

	cin>>a>>b>>c>>d;
	dp[a][b] = 1;

	for(ll i = a;i <= c;i++){
        for(ll j = b;j <= d;j++){
            if(i == a && j == b) continue;
            (dp[i][j] = (dp[i][j - 1] * i) % mod + (dp[i - 1][j] * j) % mod - (((dp[i - 1][j - 1] * (i - 1) ) % mod )* (j - 1)) % mod) %= mod;
         }
	}

	cout<<(dp[c][d] + mod) % mod;

	return 0;
}