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>
#define int long long
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define F first
#define S second

using namespace std;
typedef pair<int,int> P;

const int MOD=998244353;
int INF=100100100100100;
int dp[3010][3010];
signed main(){
  int a,b,c,d;
  cin>>a>>b>>c>>d;
  dp[a][b]=1;
  for(int i=a;i<=c;i++){
    for(int j=b;j<=d;j++){
      if(i>a)(dp[i][j]+=dp[i-1][j]*j)%=MOD;
      if(j>b)(dp[i][j]+=dp[i][j-1]*i)%=MOD;
      if(i>a && j>b)(dp[i][j]+=MOD-(dp[i-1][j-1]*(i-1)*(j-1))%MOD)%=MOD;
    }
  }
  cout<<dp[c][d]<<endl;
      return 0;
}