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 <iostream>
#include <set>
#include <vector>

using namespace std;

int n;
vector<int> v[110000];
vector<int> w[110000];
bool visited[110000];

set<int> stx;
set<int> sty;

long long ans;

void dfs(int i)
{
  if (visited[i]) return;
  visited[i] = true;
  int cnt = 0;
  for (int y: v[i]) {
    sty.insert(y);
    for (int x: w[y]) {
      stx.insert(x);
      dfs(x);
    }
  }
}

int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    w[y].push_back(x);
  }
  
  for (int i = 1; i <= 100000; i++) {
    dfs(i);
    ans += stx.size() * sty.size();
    stx.clear();
    sty.clear();
  }
  ans -= n;
  
  cout << ans << endl;
  return 0;
}