There are N dots in a two-dimensional plane. The coordinates of the i-th dot are (x_i, y_i).

We will repeat the following operation as long as possible:

* Choose four integers a, b, c, d (a \neq c, b \neq d) such that there are dots at exactly three of the positions (a, b), (a, d), (c, b) and (c, d), and add a dot at the remaining position.



We can prove that we can only do this operation a finite number of times. Find the maximum number of times we can do the operation.

Constraints

* 1 \leq N \leq 10^5
* 1 \leq x_i, y_i \leq 10^5
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
x_1 y_1
:
x_N y_N


Output

Print the maximum number of times we can do the operation.

Examples

Input

3
1 1
5 1
5 5


Output

1


Input

2
10 10
20 20


Output

0


Input

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5


Output

16
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef long long ll;
const int MOD = 1000000007;
vector<vector<int>>G(200200,vector<int>());
vector<bool>used(200200,false);
const int MX=100000;
ll xx,yy,e;
void dfs(int v){
  if(v<MX){xx++;}
  else{yy++;}
  e+=G[v].size();
  for(int nx:G[v]){
    if(used[nx])continue;
    used[nx]=true;
    dfs(nx);
  }
}
int main()
{
  int N;
  cin >> N;
  rep(i,N){
    int x,y;
    cin >> x >> y;
    x--;y--;y+=MX;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  ll ans=0;
  rep(i,200200){
    if(!used[i]){
      used[i]=true;
      xx=yy=e=0;
      dfs(i);
      e/=2;
      ans+=xx*yy-e;
    }
  }
  cout << ans << endl;


  return 0;
}