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"
using namespace std;
#define REP(i, n) for(ll i = 0;i < n;i++)
#define ll long long
#define MOD 1000000007

vector<vector<pair<int, int>>> g;
const int N = 200002;
const int N1 = 100001;
int used[N];
int col[N];

void dfs(int i, int c) {
  if(used[i]) return;
  used[i] = 1;
  col[i] = c;
  for(auto to : g[i]) {
    dfs(to.first, c);
  }
}



int main(){
  int n;
  cin>>n;
  g.resize(N);
  for(int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--; y--;
    g[x].emplace_back(y+N1, 0);
    g[y+N1].emplace_back(x, 0);
  }

  REP(i,N1){
    if (col[i]==0){
      if (g[i].size()!=0)dfs(i, i+1);
    }
  }



  vector<pair<ll, ll> > pairs(N1);
  REP(i,N1){
    if (col[i]!=0)pairs[col[i]-1].first++;
    if (col[i+N1]!=0)pairs[col[i+N1]-1].second++;
  }

  ll ans=0LL;
  REP(i,N1){
    auto to=pairs[i];
    ans+=to.first*to.second;
  }


cout<<ans-n<<endl;
  
  return 0;
}