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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
int mx = 1e5;
class disjoint_set {
  int n;
  vector<int> prt;
  public:
    vector<int> hor;
    vector<int> ver;
    disjoint_set (int n)
      : n(n), prt(n, -1), hor(n, 0), ver(n, 0)
      {
        for (int i = 0; i < n; i++) {
          for (int i = 0; i < mx; i++) {
            hor[i] = 1;
            ver[mx + i] = 1;
          }
        }
      }
      inline bool is_root (int x) const {
        return prt[x] < 0;
      }
      inline int size (int x) {
        return -prt[find(x)];
      }
      inline bool equiv (int x, int y) {
        return find(x) == find(y);
      }
      int find (int x) {
        return (is_root(x) ? x : (prt[x] = find(prt[x])));
      }
      inline bool unite (int x, int y) {
        x = find(x); y = find(y);
        if (x == y) return false;
        if (size(x) < size(y)) swap(x, y);
        prt[x] += prt[y];
        hor[x] += hor[y];
        ver[x] += ver[y];
        prt[y] = x;
        return true;
      }
};
int main() {
  cin.tie(0); cin.sync_with_stdio(false);
  int n;
  cin >> n;
  disjoint_set djs(2 * mx);
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    djs.unite(x, mx + y);
  }
  long long ret = 0;
  for (int i = 0; i < 2 * mx; i++) {
    if (!djs.is_root(i)) continue;
    if (djs.hor[i] == 0) continue;
    if (djs.ver[i] == 0) continue;
    ret += (long long)djs.hor[i] * djs.ver[i];
  }
  ret -= n;
  cout << ret << endl;
  return 0;
}