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 (ll i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll INF=1e18;
vector<ll> x[210005];
vector<ll> used(210005,0);
vector<ll> c(2);
ll cx,cy;

void dfs(ll t){
  //used[n][t]=1;
  if (t>=100005)cx++;
  if (t<100005)cy++;
  used[t]=1;
  for(ll to:x[t]){
    if (used[to])continue;
    dfs(to);
  }
}


int main() {
  ll N;
  cin >> N;
  vector<P> v(N);


  rep(i,N){
    ll a1,b1;
    cin>>a1>>b1;
    a1--;b1+=100005;
    v[i]=make_pair(a1,b1);
    x[a1].push_back(b1);
    x[b1].push_back(a1);
  }
ll ans=0;
  rep(i,N){
    c[0]=0;c[1]=0;
cx=0;cy=0;
    dfs(v[i].first);
  //  cout<<"i"<<i<<cx<<" "<<cy<<endl;
    ans+=cx*cy;

  }

  ans-=N;
  cout<<ans;

}
