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() {
  int A,B,C,D;
  cin>>A>>B>>C>>D;
  int64_t Q=998244353;
  vector<vector<int64_t>> dp(C+2,vector<int64_t>(D+2));
  dp.at(A).at(B)=1;
  for(int i=A;i<C;i++){
    dp.at(i+1).at(B)=(dp.at(i).at(B)*B)%Q;
  }
  for(int i=B;i<D;i++){
    dp.at(A).at(i+1)=(dp.at(A).at(i)*A)%Q;
  }
  for(int i=A;i<C;i++){
    for(int j=B;j<D;j++){
      dp.at(i+1).at(j+1)=(dp.at(i).at(j+1)*(j+1)+dp.at(i+1).at(j)*(i+1)-dp.at(i).at(j)*i*j+Q*i*j)%Q;
    }
  }
  cout<<dp.at(C).at(D)<<endl;    
}