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;
#define ll long long
#define fastIO cout << fixed << setprecision(12), ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
double const EPS = 1e-10, PI = acos(-1);
const int N = 2e5 + 9, M = 1e5 + 9, OO = 1e9 + 7, MOD = 1e9 + 7;
const ll inf = 1e18;

vector<int> adj[N];
bool vis[N];
ll Xs, Ys;

void dfs(int node) {
  if(node > 1e5)
    ++Ys;
  else
    ++Xs;
  vis[node] = 1;
  for(int it: adj[node]) {
    if(vis[it]) continue;
    dfs(it);
  }
}

int main() {
  fastIO;
#ifdef LOCAL
  freopen("input.in", "rt", stdin);
#endif
  int n, x, y;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> x >> y;
    y += 1e5;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  ll total = 0;
  for (int i = 0; i < N; ++i) {
    if(!vis[i]) {
      Xs = Ys = 0;
      dfs(i);
      total += (Xs * Ys);
    }
  }
  cout << total - n;
  return 0;
}
