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;
typedef long long ll;
typedef pair<int,int> P;
#define rep(i,m,n) for(int i=m;i<n;i++)
ll mod=998244353;

int main(){
  int a,b,c,d;
  cin>>a>>b>>c>>d;
  ll dp[c+1][d+1];
  rep(i,0,c+1){
    rep(j,0,d+1){
      dp[i][j]=0;
    }
  }
  dp[a][b]=1;
  rep(i,b,d){
    dp[a][i+1]=dp[a][i]*a;
    dp[a][i+1]%=mod;
  }
  rep(i,a,c){
    dp[i+1][b]=dp[i][b]*b;
    dp[i+1][b]%=mod;
  }
  rep(i,a,c){
    rep(j,b,d){
      dp[i+1][j+1]=dp[i][j+1]*(j+1)%mod+dp[i+1][j]*(i+1)%mod-dp[i][j]*i%mod*j%mod;
      if(dp[i+1][j+1]<0) dp[i+1][j+1]+=mod;
      else dp[i+1][j+1]%=mod;
    }
  }
  cout<<dp[c][d]<<endl;
}