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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
vector<int> g[202020];
int done[202020];
pair<int, int> dfs(int cur) {
  int u = 0, v = 0;
  for (int nbr : g[cur]) {
    if (done[nbr]) continue;
    done[nbr] = true;
    pair<int, int> res = dfs(nbr);
    u += res.first;
    v += res.second;
  }
  if (cur < 100000) u++;
  else v++;
  return make_pair(u, v);
}
signed main() { 
  ios::sync_with_stdio(false); cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y + 100000);
    g[y + 100000].push_back(x);
  }
  long long ans = 0;
  for (int i = 0; i < 202020; i++) {
    if (done[i]) continue;
    done[i] = true;
    pair<int, int> res = dfs(i);
    ans += 1LL * res.first * res.second;
  }
  cout << ans - n << endl;
  return 0;
}