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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include<unordered_map>
#include<utility>
template<typename T>
class UnionFind {
 public:
  void unite(const T& a, const T& b) {
    auto x = find(a), y = find(b);
    if(rank_[x] < rank_[y]) std::swap(x, y);
    parent_[y] = x;
    if(rank_[x] == rank_[y]) rank_[x] += 1;
  }
  T find(const T& x) {
    if(!parent_.count(x)) {
      rank_[x] = 1;
      return parent_[x] = x;
    }
    return (parent_[x] == x) ? x : (parent_[x] = find(parent_[x]));
  }
 private:
  std::unordered_map<T, T> parent_;
  std::unordered_map<T, int> rank_;
};

#include<bits/stdc++.h>
using namespace std;

struct Point {int id, x, y;};

using LL = long long;

int main() {
  int N;
  cin >> N;
  vector<Point> P;
  for(auto i=0; i<N; ++i) {
    int x, y;
    cin >> x >> y;
    P.push_back({i, x, y});
  }

  UnionFind<int> uf;
  sort(begin(P), end(P), [](auto l, auto r) {return l.x < r.x;});
  for(auto i=1; i<N; ++i) if(P[i-1].x == P[i].x) uf.unite(P[i-1].id, P[i].id);
  sort(begin(P), end(P), [](auto l, auto r) {return l.y < r.y;});
  for(auto i=1; i<N; ++i) if(P[i-1].y == P[i].y) uf.unite(P[i-1].id, P[i].id);

  set<int> G;
  map<int, set<int>> X, Y;
  for(const auto& p: P) {
    auto g = uf.find(p.id);
    G.emplace(g);
    X[g].emplace(p.x);
    Y[g].emplace(p.y);
  }

  LL sum = 0;
  for(auto g: G) {
    LL x = X[g].size() - 1;
    LL y = Y[g].size() - 1;
    sum += x * y;
  }
  cout << sum << endl;
}