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;
using ll = long long;
const int N = 1e5;
void dfs(ll node,vector<vector<ll>>& adj,vector<bool>& vu,set<ll>& s){
  if(vu[node]) return;
  s.insert(node);
  vu[node] = true;
  for(ll i : adj[node]) dfs(i,adj,vu,s);
}
int main(){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n,x,y;
  cin >> n;
  vector<vector<ll>> adj(2*N);
  for(int i = 0; i < n; ++i){
    cin >> x >> y;
    adj[x-1].push_back(N+y-1);
    adj[N+y-1].push_back(x-1);
  }
  ll ret = -n;
  vector<bool> vu(2*N);
  for(ll i = 0; i < 2*N; ++i){
    set<ll> s;
    dfs(i,adj,vu,s);
    ll nbx = 0,nby = 0;
    for(ll j : s){
      if(j < N) ++nbx;
      else ++nby;
    }
    ret += nbx*nby;
  }
  cout << ret;
  return 0;
}
