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 <iostream>
#include <vector>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

using Graph = vector<vector<int>>;

// int の最大値は2147483647
// xとyの最大値は1.0e5だから。
const int maxX = 100005;

vector<bool> seen(maxX*2);

vector<int> nc(2);

void dfs(Graph& _G, int _v){
  seen[_v] = true;
  if(_v < maxX) nc[0] += 1;
  else nc[1] += 1;

  for(auto nv: _G[_v]){
    if(seen[nv]) continue;
    dfs(_G, nv);
  }
}

int main(){
  int N;
  cin >> N;

  Graph G(maxX*2);

  for(int i = 0 ; i < N; i++){
    int x,y;
    cin >> x >> y;
    y+=maxX;
    G[x].push_back(y);
    G[y].push_back(x);
  }

  ll counter = 0;
  seen.assign(maxX*2, false);

  for(int i = 0; i < maxX*2; i++){  //Graphのすべてのノードについて
    if(seen[i]) continue;
    nc.assign(2, 0);
    dfs(G, i);
    counter += (ll)nc[0] * nc[1]; //二部グラフのノード数を掛け合わせれば、辺の数がわかる
  }
  counter -= N;

  cout << counter << endl;

  return 0;
}
