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;

#define int long long
const int MOD=998244353;
int dp[4001][4001];
int f(int a,int b){
  if(~dp[a][b])return dp[a][b];
  return dp[a][b]=(f(a-1,b)*b%MOD+f(a,b-1)*a%MOD+MOD-f(a-1,b-1)*(a-1)%MOD*(b-1)%MOD)%MOD;
}

signed main(){
  int a,b,c,d;cin>>a>>b>>c>>d;
  memset(dp,-1,sizeof(dp));
  for(int i=0;i<=4000;i++)for(int j=0;j<=4000;j++)if(i<a||j<b)dp[i][j]=0;
  dp[a][b]=1;
  //for(int i=0;i<10;i++)for(int j=0;j<10;j++)cout<<i<<" "<<j<<" "<<f(i,j)<<endl;
  cout<<f(c,d)<<endl;
}