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

class unionfind {
  vector<int> parent;
  public:  
  void init(int n) {
    for (int i = 0; i < n; i++) {
      parent.push_back(i);
    }
  }
  
  void merge(int m, int n) {
    if (find(m) == find(n)) return;
    parent.at(find(m)) = find(n);
  }
  
  int find(int n) {
    if (parent.at(n) == n) return n;
    return parent.at(n) = find(parent.at(n));
  }
};

int main()
{
  int n;
  cin >> n;
  unionfind t;
  t.init(200000);
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    t.merge(x - 1, y + 99999);
  }
  vector<int> sx(200000), sy(200000);
  for (int i = 0; i < 100000; i++) {
    sx.at(t.find(i))++;
    sy.at(t.find(i + 100000))++;
  }
  long s = 0;
  for (int i = 0; i < 200000; i++) {
    s += (long)sx.at(i) * sy.at(i);
  }
  cout << s - n << endl;
}
