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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<unordered_set>
#include<unordered_map>
using namespace std;
typedef long long ll;
#define mod 1000000007
#define mad(a,b) a=(a+b)%mod;
#define N 100010
vector<ll> g[N*2];
ll n;
bool vis[N*2];
ll cnt1,cnt2;
void dfs(ll x){
    if(vis[x])return; vis[x]=1;
    if(x<N)cnt1++; else cnt2++;
    for(auto y:g[x])dfs(y);
}
int main(){
  cin>>n;
  for(int i=0;i<n;i++){
      ll x,y;cin>>x>>y;
      y+=N;
      g[x].push_back(y);
      g[y].push_back(x);
  }
  for(int i=0;i<2*N;i++)vis[i]=0;
  ll ans=-n;
  for(int i=1;i<N;i++){
      cnt1=0,cnt2=0;
      dfs(i);
      //if(cnt1>0||cnt2>0)cout<<cnt1<<" "<<cnt2<<endl;
      ans+=cnt1*cnt2;
  }
  cout<<ans<<endl;
  return 0;
}
