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>
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

vector<vector<int>> g;
vector<bool> chk;
vector<ll> cnt;

void dfs(int p)
{
  if (chk[p])
    return;
  chk[p] = true;
  cnt[p%2]++;
  for (int i : g[p])
    dfs(i);
}

int main()
{
  int n;
  cin >> n;
  g.resize(200001);
  rep(i, n)
  {
    int x, y;
    cin >> x >> y;
    x = x * 2 - 2;
    y = y * 2 - 1;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  ll ans = 0;
  chk.resize(200001);
  fill(chk.begin(), chk.end(), false);
  cnt.resize(2);
  rep(i, 200001)
  {
    fill(cnt.begin(), cnt.end(), 0);
    dfs(i);
    ans += cnt[0] * cnt[1];
  }
  cout << ans - n << endl;
  return 0;
}