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;

#undef LOCAL
#ifdef LOCAL
#include "../lib/dump.hpp"
#else
#define dump(...)
#define dumpv(...)
#endif

typedef long long ll;
const ll MOD = 1e9 + 7;
const ll INF = 1e18;

#define VMAX 100001
// #define VMAX 50
vector<vector<int>> to(VMAX * 2);
vector<int> used(VMAX * 2);
int mx = 0;
int my = 0;

void dfs(int v)
{
  if (used[v])
    return;
  used[v] = true;
  if (v < VMAX)
    mx++;
  else
    my++;
  dump(v, mx, my);
  dumpv(to[v]);
  for (int u : to[v]) {
    dfs(u);
  }
}

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  // freopen("temp.1", "r", stdin);

  int N;
  cin >> N;
  for (int i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;
    y += VMAX;
    to[x].push_back(y);
    to[y].push_back(x);
  }
  dumpv(used);
  ll ans = 0;
  for (int i = 0; i < VMAX * 2; i++) {
    mx = 0, my = 0;
    dfs(i);
    ans += (ll)mx * my;
  }
  dumpv(used);
  cout << ans - N << endl;
  return 0;
}
