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 int long long

#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif

int V = 101010;
vector<int> graph[101010*2];

signed main(){

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

  vector<int> used(V * 2,false);
  vector<int> cnt(2,0);
  function< void(int) > dfs =
  [&](int u){
    if(used[u]) return;
    used[u] = true;
    cnt[u/V]++;
    for(int v : graph[u]){
      dfs(v);
    }
  };

  int ans = 0;
  for(int i = 0; i < V*2; i++){
    cnt = vector<int>(3,0);
    dfs(i);
    ans += cnt[0] * cnt[1];
  }
  cout << ans - n << endl;

  return 0;
}
