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;

using ll=long long;
const ll MOD=998244353;
ll dp[4001][4001],a,b,c,d;
ll f(ll A,ll B){
  if(A<a||B<b)return 0;
  if(~dp[A][B])return dp[A][B];
  return dp[A][B]=(MOD+f(A-1,B)*B+f(A,B-1)*A-f(A-1,B-1)*(A-1)*(B-1)%MOD)%MOD;
}

int main(){
  cin>>a>>b>>c>>d;
  memset(dp,-1,sizeof(dp));
  dp[a][b]=1;
  cout<<f(c,d)<<endl;
}