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<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll dp[3001][3001],a,b,c,d,MOD=998244353;
int main(void){
    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-1][j]*j%MOD+dp[i][j-1]*i%MOD;
            dp[i][j]%=MOD;
            if(i!=a&&j!=b)dp[i][j]-=dp[i-1][j-1]*(i-1)%MOD*(j-1)%MOD;
            dp[i][j]%=MOD;
        }
    }
    //cout<<dp[2][3]<<" "<<dp[3][3]<<" "<<dp[3][2]<<endl;
    cout<<(MOD+dp[c][d])%MOD<<endl;
}