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;
typedef long long ll;
typedef pair<ll,ll> P;


vector<ll> v[210005];
vector<ll> cnt(2);
vector<ll> used(210005,0);



int dfs(ll x){
  if(x>=100005)cnt[1]++;
  if(x<100005)cnt[0]++;
  used[x]=1;
  for(ll t:v[x]){
    if (used[t])continue;
    dfs(t);
  }
  return 0;
}
int main() {
  ll n;
  cin>>n;
  vector<P> p;
  rep(i,n){
    ll x1,y1;
    cin>>x1>>y1;
    x1--; y1+=100005;
    v[x1].emplace_back(y1);
    v[y1].emplace_back(x1);
    p.push_back(make_pair(x1,y1));
  }
  ll ans=0;
  rep(i,n){
    dfs(p[i].first);
    ans+=cnt[0]*cnt[1];
    cnt[0]=0;cnt[1]=0;
  }
  ans-=n;
  cout<<ans;
}
