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;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=1LL<<61;

const ll N=2e5+10;
vector<set<ll>> G(N);
vec al(N);
ll x,y;
void dfs(ll f) {
  al[f]=1;
  if(f<1e5)x++;
  else y++;
  for(ll t:G[f]) {
    if(!al[t]) dfs(t);
  }
}

int main() {
  ll n;
  cin >> n;
  for(ll i=0;i<n;i++) {
    ll x,y;
    cin >> x >> y;
    x--;y--;
    y+=1e5;
    G[x].insert(y);
    G[y].insert(x);
  }
  ll ans=0;
  for(ll i=0;i<2e5;i++) {
    if(!al[i]&&G[i].size()) {
      x=0,y=0;
      dfs(i);
      ans+=x*y;
    }
  }
  cout << ans-n << endl;
}