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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main()
{
  int n;
  cin >> n;
  vector<pair<int, int>> p(n), q(n);
  for (int i = 0; i < n; i++) {
    cin >> p.at(i).first >> p.at(i).second;
    q.at(i) = make_pair(p.at(i).second, p.at(i).first);
  }
  sort(p.begin(), p.end());
  sort(q.begin(), q.end());
  set<int> activex, activey;
  pair<int, int> prev;
  for (int i = 0; i < n; i++) {
    if (p.at(i).first == prev.first) {
      activey.insert(prev.second);
      activey.insert(p.at(i).second);
    }
    prev = p.at(i);
  }
  prev = make_pair(0, 0);
  for (int i = 0; i < n; i++) {
    if (q.at(i).first == prev.first) {
      activex.insert(prev.second);
      activex.insert(q.at(i).second);
    }
    prev = q.at(i);
  }
  long actives = 0;
  for (int i = 0; i < n; i++) {
    if (activex.find(p.at(i).first) != activex.end()
        && activey.find(p.at(i).second) != activey.end()) {
      actives++;
    }
  }
  cout << activex.size() * activey.size() - actives << endl;
}