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 fi first
#define endl "\n"
#define se second
#define ll long long
#define inf 0x3f3f3f3f
#define fast    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
const int N = 2e5+5;
const ll mod =   998244353;
ll a,b,c,d;
ll dp[3050][3050];
ll fn(ll x, ll y) {
	if(x<a||y<b)
	return 0;
	if(x==a&&y==b)
	return 1;
	if(dp[x][y]!=-1)
	return dp[x][y];
	return dp[x][y]=(x*fn(x,y-1)+y*fn(x-1,y)-(x-1)*(y-1)*fn(x-1,y-1)%mod+mod)%mod;
}
int main()
{  fast
cin>>a>>b>>c>>d;
memset(dp , -1, sizeof dp);
cout<<fn(c,d);
}