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>

#define int long long
#define F first
#define S second
#define pb push_back
#define pii pair <int,int>

using namespace std;

const int N=3309,MOD=998244353;
int dp[N][N];

int32_t main(){
    ios::sync_with_stdio(0);cin.tie(0);
    int a,b,c,d;
    cin>>a>>b>>c>>d;
    dp[a][b]=1;
    for(int i=a;i<=c;i++){
        for(int j=b;j<=d;j++){
            if(i==a && j==b) continue ;
            dp[i][j]=(dp[i][j-1]*i+dp[i-1][j]*j-dp[i-1][j-1]*(i-1)*(j-1)+MOD*MOD)%MOD;
        }
    }
    cout<<dp[c][d];
    return 0;
}