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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; i++)

const int G = 2e5;
vector<int> e[G+1];
bool used[G+1];

const int yBase = 1e5;

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

  rep(i,N) {
    int x, y;
    cin >> x >> y;
    e[x].push_back(yBase+y);
    e[yBase+y].push_back(x);
  }

  ll ans = 0;
  for(int i = 1; i <= G; i++) {
    if(used[i]) continue;

    ll nx, ny, ne;
    nx = ny = ne = 0;
    queue<int> q;
    q.push(i);
    while(!q.empty()) {
      int u = q.front(); q.pop();
      if(used[u]) continue;
      used[u] = true;
      if(u <= yBase) {
        nx++;
      } else {
        ny++;
      }

      for(auto v: e[u]) {
        ne++;
        if(used[v]) continue;
        q.push(v);
      }
    }
    if(nx >= 2 && ny >= 2) ans += nx * ny - ne/2;
  }

  cout << ans << endl;

  return 0;
}