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;

int main(){
  long long a,b,c,d,i,j;
  cin >> a >> b >> c >> d;
  vector<vector<long long>> dp(c+1,vector<long long>(d+1));
  long long p=998244353;
  
  dp.at(a).at(b)=1;
  for(i=a+1;i<=c;i++){
    dp.at(i).at(b)=(dp.at(i-1).at(b)*b)%p;
  }
  for(j=b+1;j<=d;j++){
    dp.at(a).at(j)=(dp.at(a).at(j-1)*a)%p;
  }
  
  for(i=a+1;i<=c;i++){
    for(j=b+1;j<=d;j++){
      dp.at(i).at(j)=(dp.at(i-1).at(j)*j+dp.at(i).at(j-1)*i-dp.at(i-1).at(j-1)*(i-1)*(j-1))%p;
    }
  }
  
  if(dp.at(c).at(d)<0){
    dp.at(c).at(d) += p;
  }
  
  cout << dp.at(c).at(d) << endl;
}