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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_set>
#include <vector>

using namespace std;

#define ll long long
#define P pair<int, int>
#define FOR(i, N) for (int i = 0; i < (int)N; i++)
#define FORIN(i, a, b) for (int i = a; i < (int)b; i++)
#define ALL(x) (x).begin(), (x).end()
#define GI(name) \
  int(name);     \
  scanf("%d", &(name))
#define GI2(name1, name2) \
  int(name1), (name2);    \
  scanf("%d %d", &(name1), &(name2))
#define GI3(name1, name2, name3) \
  int(name1), (name2), (name3);  \
  scanf("%d %d %d", &(name1), &(name2), &(name3))
#define GVI(name, size)    \
  vector<int>(name)(size); \
  FOR(i, (size)) scanf("%d", &(name)[i])
#define GS(name) \
  string(name);  \
  cin >> (name);
#define MOD 1000000007
#define DEBUG(...) debug(__LINE__, ":" __VA_ARGS__)

string to_string(string s) {
  return s;
}

template <class T>
string to_string(vector<T> v) {
  string ret = "{";
  FOR(i, v.size() - 1) { ret += to_string(v[i]) + ","; }
  if (v.size() > 0) {
    ret += to_string(v.back());
  }
  ret += "}";
  return ret;
}

void debug() {
  cerr << endl;
}

template <class Head, class... Tail>
void debug(Head head, Tail... tail) {
  cerr << to_string(head) << " ";
  debug(tail...);
}

void print() {
  cout << endl;
}

template <class Head, class... Tail>
void print(Head head, Tail... tail) {
  cout << to_string(head);
  print(tail...);
}

struct Union_Find {
  vector<int> data;
  Union_Find(int size) { data = vector<int>(size, -1); }
  bool union_set(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (data[y] < data[x])
        swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
    return x != y;
  }
  bool find(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  int size(int x) { return -data[root(x)]; }
};

int main() {
  GI(N);
  int siz = 10;
  Union_Find ufx(siz), ufy(siz);
  vector<vector<int>> listx(siz), listy(siz);
  FOR(i, N) {
    GI2(x, y);
    if (listy[y].size() > 0) {
      ufx.union_set(x, listy[y][0]);
    }
    if (listx[x].size() > 0) {
      ufy.union_set(y, listx[x][0]);
    }
    listx[x].push_back(y);
    listy[y].push_back(x);
    debug(ufx.data, ufy.data);
  }
  vector<unordered_set<int>> unions(siz);
  vector<int> sizes(siz);
  FOR(i, siz) {
    if (ufx.root(i) >= 0) {
      for (int y : listx[i]) {
        unions[ufx.root(i)].insert(y);
      }
      sizes[ufx.root(i)] += listx[i].size();
    } else if (ufx.data[i] < -1) {
      for (int y : listx[i]) {
        unions[i].insert(y);
      }
      sizes[i] += listx[i].size();
    }
  }
  ll ans = 0;
  FOR(i, siz) {
    if (ufx.data[i] < -1) {
      ans += -ufx.data[i] * unions[i].size();
      ans -= sizes[i];
    }
  }
  print(ans);
  return 0;
}