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;
typedef long long ll;
#define rep(i,m,n) for(int i=m;i<n;i++)

bool used[100000*2];
vector<int> a[100000*2];
int x,y;

void dfs(int v){
  if(used[v]) return;
  used[v]=true;
  if(v<100000) x++;
  else y++;
  for(int u:a[v]){
    dfs(u);
  }
}

int main(){
  int n;
  cin>>n;
  rep(i,0,n){
    int x,y;
    cin>>x>>y;
    x--;
    y--;
    a[x].push_back(y+100000);
    a[y+100000].push_back(x);
  }
  rep(i,0,100000*2) used[i]=false;
  ll ans=0;
  rep(i,0,100000*2){
    x=0;
    y=0;
    dfs(i);
    ans+=(ll)x*y;
  }
  cout<<ans-n<<endl;
}