Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

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

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
using namespace std;
typedef long long ll;
struct UFT{
  vector<int> par;//親
  vector<int> rank;//木の深さ
  vector<int> size;//木の大きさ
  int n;
  
  UFT(int _n)
  {
    n = _n;
    par.resize(n);
    rank.assign(n,0);
    size.assign(n,0);
    rep(i,n){
      par[i] = i;
    }
  }
  
  //xの根を返す
  int find(int x)
  {
    if(par[x] == x) return x;
    else return par[x] = find(par[x]);
  }

  //x,yを併合
  void unite(int x,int y)
  {
    x = find(x);
    y = find(y);
    if(x == y) return;

    if(rank[x] < rank[y]){
      par[x] = y;
      size[y] += size[x];
    }
    else{
      par[y] = x;
      size[x] += size[y];
      if(rank[x] == rank[y]) rank[x]++;
    }
  }

  //x,yが同じグループにいるかどうかを返す
  bool same(int x,int y)
  {
    return find(x) == find(y);
  }

  //xの属する木のサイズを探す
  int usize(int x)
  {
    return size[find(x)];
  }
};
const int M = 100010;
int main()
{
  ll n;cin >> n;
  vector<int> x(n), y(n);
  rep(i,n) {
    cin >> x[i] >> y[i];
    x[i]--; y[i]--;
  }

  UFT uf(2*M);
  rep(i,n) uf.unite(x[i], y[i] + M);
  map<int,ll> mx, my;
  rep(i, M) mx[ uf.find(i) ]++;
  rep(i, M) my[ uf.find(i + M) ]++;
  
  ll res = 0;
  rep(i, 2*M) res += mx[i]*my[i];
  cout << res - n << "\n";
  
  return 0;
}