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)
#define sz(x) int(x.size())
#define show(x) {for(auto i: x){cout << i << " ";} cout << endl;}
using namespace std;
using ll = long long;
using P = pair<int, int>;

const int V = 100005;
vector<int> to[V*2];
bool visited[V*2];

vector<int> cnt;
void dfs(int x) {
  if (visited[x]) return;
  visited[x] = true;
  cnt[x/V]++;
  for (int y : to[x]) {
    if (!visited[y]) dfs(y);
  }
}

int main() {
  // yに+100005、to[]にx→yを入れる、同じグループのx座標の数とy座標の数の積
  int N;
  cin >> N;
  rep(i, N) {
    int X, Y;
    cin >> X >> Y;
    Y += V;
    to[X].push_back(Y);
    to[Y].push_back(X);
  }
  ll ans = 0;
  for (int i = 1; i < V+1; i++) {
    if (visited[i]) continue;
    cnt = vector<int> (2);
    dfs(i);
    ans += (ll)cnt[0] * cnt[1];
  }
  ans -= N;
  cout << ans << '\n';
  return 0;
}