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;

int64_t cnt_x;
int64_t cnt_y;
const int vn = 100000;
vector<vector<int>> to(vn * 2);
vector<bool> visited(vn * 2, false);

void dfs(int v)
{
  if (visited[v]) return;
  visited[v] = true;
  if (v < vn) {
    cnt_x++;
  } else {
    cnt_y++;
  }
  for (auto&& nv : to[v]) {
    dfs(nv);
  }
}

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

  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    y += vn;
    to[x].push_back(y);
    to[y].push_back(x);
  }

  int64_t ans = 0;
  for (int i = 0; i < vn * 2; i++) {
    if (visited[i]) continue;
    cnt_x = 0;
    cnt_y = 0;
    dfs(i);
    ans += cnt_x * cnt_y;
  }
  cout << ans - n << endl;
}
