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 <vector>
using namespace std;

bool xy[200010];
typedef long long ll;

vector<int> g[200010];

ll XX;
ll YY;
ll ans;

void dfs(int v) {
  xy[v] = true;

  if(v < 100000) XX++;
  else YY++;

  for(int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    if(xy[u]) continue;
    dfs(u);
  }
  return;
}
    
int main() {
  int n;
  cin >> n;

  for(int i = 0; i < n; i++) {
    int X,Y;
    cin >> X >> Y;
    X--;
    Y--;
    Y += 100000;
    g[X].push_back(Y);
    g[Y].push_back(X);
  }

  for(int i = 0; i < 200000; i++) {
    if(xy[i]) continue;
    XX = 0;
    YY = 0;
    dfs(i);
    ans += XX * YY;
  }

  cout << ans - n << endl;

}
