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;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; i++)

const int G = 2e5;
vector<int> e[G+1];
bool used[G+1];

const int yBase = 1e5;

ll nx, ny;

void dfs(int u) {
  if(used[u]) return;
  used[u] = true;
  if(u <= yBase) {
    nx++;
  } else {
    ny++;
  }

  for(auto v: e[u]) dfs(v);
}

int main () {
  int N;
  cin >> N;

  rep(i,N) {
    int x, y;
    cin >> x >> y;
    e[x].push_back(yBase+y);
    e[yBase+y].push_back(x);
  }

  ll ans = -N;
  for(int i = 1; i <= G; i++) {
    if(used[i]) continue;
    nx = ny = 0;
    dfs(i);
    ans += nx * ny;
  }

  cout << ans << endl;

  return 0;
}
