Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in C++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#define vv(a, b, c, d) vector<vector<d> >(a, vector<d>(b, c))
#define vvi std::vector<std::vector<int> >
#define vvl std::vector<std::vector<ll> >
#define MODs 1000000007;
typedef long long int ll;
using namespace std;
vvl G = vv(200002, 0, 0, ll);
std::vector<ll> check(200002, 0);
ll ans = 0;
set<ll> x, y;
void dfs(ll from, ll now){
  if(check[now]!=0) return;
  check[now] = 1;
  if(now>=100001&&y.find(now)==y.end()) y.insert(now);
  if(now<100001&&x.find(now)==x.end()) x.insert(now);
  for(int i=0;i<G[now].size();i++){
    if(from==G[now][i]) continue;
    dfs(now, G[now][i]);
  }
}
int main(int argc, char const *argv[]) {
  ll N, a, b, ans = 0;
  std::cin >> N;
  for(int i=0;i<N;i++){
    std::cin >> a >> b;
    G[a].push_back(b+100001);
    G[b+100001].push_back(a);
  }
  for(int i=1;i<200002;i++){
    x.clear(), y.clear();
    dfs(-1, i);
    ans += x.size()*y.size();
  }
  std::cout << ans - N << '\n';
  return 0;
}