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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#define rep(i,x,y) for(ll i=(x); i<(y); i++)

using namespace std;
typedef long long int ll;
typedef pair<ll,ll> PLL;

const ll V = 1000000; // 片方の頂点の最大数
vector<ll> to[2 * V]; // xあるいはyからつながっている先。y側はVの下駄を履かせる
bool visited[2 * V];
vector<ll> cnt;

void dfs(int v) {
  if (visited[v])
    return;
  visited[v] = true;
  cnt[v/V]++;
  for (ll u : to[v]) {
    dfs(u);
  }
}

signed main(void) {
  // (x,y)座標を二部グラフにする
  ll N;
  cin>>N;
  rep(i,0,N) {
    ll x, y;
    cin>>x>>y;
    y += V;
    to[x].push_back(y);
    to[y].push_back(x);
  }

  ll ans = 0;
  rep(i,0,2*V){
    if (visited[i])
      continue;
    cnt = vector<ll>(2);
    dfs(i);
    ans += cnt[0]*cnt[1];
  } 
  ans -= N;
  cout<<ans<<endl;
  return 0;
}
