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 <algorithm>
#include <cassert>
#include <cstring>
#include <iostream>
#include <vector>
#include <set>
using namespace std;

typedef long long ll;

struct Pos { int x, y; };

const int v = 100000;
bool used[v*2];
vector<int> to[v*2];

void dfs(int s, set<int> &sx, set<int> &sy) {
  ll ret = 0;
  used[s] = true;
  if (s >= v) sy.insert(s);
  else sx.insert(s);
  for (int t: to[s]) {
    if (used[t]) continue;
    dfs(t, sx, sy);
  }
  return;
}

int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) { 
    int x, y;
    cin >> x >> y;
    --x; --y;
    y += v;
    to[x].push_back(y);
    to[y].push_back(x);
  }
  ll ans = 0;
  for (int i = 0; i < v*2; ++i) {
    if (used[i]) continue;
    set<int> sx, sy;
    dfs(i, sx, sy);
    ans += (ll)sx.size()*(ll)sy.size();
  }
  ans -= n;
  cout << ans << endl;
}
