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;
  vector<vector<int64_t>> x(c-a+1,vector<int64_t>(d-b+1));
  int64_t p=998244353;
  x[0][0]=1;
  for(int i=0;i<=c-a;i++){
    for(int j=0;j<=d-b;j++){
      if(i==0&&j!=0)
        x[i][j]=x[i][j-1]*a%p;
      if(i!=0&&j==0)
        x[i][j]=x[i-1][j]*b%p;
      if(i!=0&&j!=0)
        x[i][j]=(x[i][j-1]*(a+i)+x[i-1][j]*(b+j)-x[i-1][j-1]*(a+i-1)*(b+j-1)+p*c*d)%p;
    }
  }
  cout<<x[c-a][d-b]<<endl;
}