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 <math.h>
#include <algorithm>
using namespace std;
long long dp[3001][3001];
int main(){
  int A,B,C,D;
  cin>>A>>B>>C>>D;
  for(int i=A;i<=C;i++){
    for(int j=B;j<=D;j++){
      if(i==A&&j==B) dp[A][B]=1;
      else if(i==A&&j!=B) dp[A][j]=(A*dp[A][j-1])%998244353;
      else if(i!=A&&j==B) dp[i][B]=(B*dp[i-1][B])%998244353;
      else{
       dp[i][j]=(i*dp[i][j-1]+j*dp[i-1][j]-((i-1)*(j-1)*(dp[i-1][j-1]%998244353))%998244353);
       if(dp[i][j]<0) dp[i][j]+=998244353;
       dp[i][j]=dp[i][j]%998244353;
      }
    }
  }
  cout<<dp[C][D]<<endl;
  
  return 0;
}