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 ll mod=1e9+7;
int dy[]={1,-1,0,0};
int dx[]={0,0,1,-1};

const int maxn=100000;
bool visited[maxn*2]={false};
int cnt[2];
vector<int>  related[2*maxn];

void dfs(int pos){
    cnt[pos/maxn]++;
    for(auto next:related[pos]){
        if(visited[next]) continue;
        visited[next]=true;
        dfs(next);
    }
}


int main(){
  int N;
  cin>>N;
  int x[maxn],y[maxn];
  for(int i=0;i<N;i++){
    cin>>x[i]>>y[i];
    related[x[i]-1].push_back(maxn+y[i]-1);
    related[maxn+y[i]-1].push_back(x[i]-1);
  }
  ll ans=0;
  for(int i=0;i<N;i++){
    if(visited[x[i]-1]) continue;
    cnt[0]=0; cnt[1]=0;
    visited[x[i]-1]=true;
    dfs(x[i]-1);
    ans+=(ll)cnt[0]*cnt[1];
  }

  ans-=N;
  cout<<ans<<endl;

  return 0;
}