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;

#define int long long

vector<int> G[200000];
const int M = 100000;
vector<bool> used(200000, false);  
using P = pair<int, int>;
int A, B;

void dfs(int v) {
  if ( used[v] ) return;  
  used[v] = true;
  if ( v >= M ) B++;
  else A++;
  for ( int u : G[v] ) dfs(u);  
}

signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);

  int N;
  cin >> N;

  for ( int i = 0; i < N; i++ ) {
    int x, y;
    cin >> x >> y;
    x--; y--;
    G[x].push_back(M+y);
    G[M+y].push_back(x);
  }

  int ans = -N;  
  for ( int i = 0; i < 200000; i++ ) {
    if ( used[i] ) continue;
    dfs(i);
    ans += A*B;
    A = B = 0;
  }

  cout << ans << endl;
  
  return 0;
}
