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;

typedef long long ll;
const int mod=998244353;

int a,b,c,d;
ll dp[3005][3005];

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>a>>b>>c>>d;
    dp[a][b]=1;
    for(int i=a+1;i<=c;i++)dp[i][b]=dp[i-1][b]*b%mod;
    for(int i=b+1;i<=d;i++)dp[a][i]=dp[a][i-1]*a%mod;
    for(int i=a+1;i<=c;i++){
        for(int j=b+1;j<=d;j++){
            dp[i][j]=((dp[i-1][j]*j+dp[i][j-1]*i-dp[i-1][j-1]*(i-1)*(j-1))%mod+mod)%mod;
        }
    }
    cout<<dp[c][d]<<endl;

    return 0;
}