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 <map>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string>
#define rep(i,n) for(int i=0;i<n;i++)

//abbreviations
typedef long long ll;
#define vi vector<int>
#define vvi vector<vector<int>>
#define vl vector<ll>
#define vs vector<string>
#define vb vector<bool>

using namespace std;

const int vmax = 100100;
vi graph[vmax*2];
vb seen(vmax*2);
vi cnt(2,0);

void dfs(int v) {
  if(seen[v]) return;
  seen[v] = true; // v を訪問済にする
  cnt[v/vmax]++;

  // v から行ける各頂点 next_v について
  for (int next_v : graph[v]) {
    dfs(next_v); // 再帰的に探索
  }
}



int main(){
  int i = 0;
  int N = 0;
  ll ans = 0;
  cin>>N;
  rep(i,N){
    int x,y;
    cin >> x >> y;
    y += vmax;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  seen.assign(vmax*2,false);
  rep(i,vmax*2){
      if(seen[i])continue;
      cnt[0]=0;cnt[1]=0;
      dfs(i);
      ans+=(ll)cnt[0]*cnt[1];
  }
  ans-=N;
  cout<<ans<<endl;

  return 0;
}
