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 <cstdio>
#include <cstdlib>
#include <cmath>

#include <iostream>

#include <string>
#include <vector>
#include <algorithm>

using namespace std;

typedef long long ll;

const ll mod = 1000000007;

class mint{
public:
  ll x;
  mint(ll x=0) : x(x%mod){}
  mint& operator+=(const mint rhs){
    x += rhs.x;
    while(x >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint rhs){
    x -= rhs.x;
    while(x < 0) x += mod;
    return *this;
  }
  mint & operator*=(const mint rhs){
    (x *= rhs.x) %= mod;
    return *this;
  }
  mint operator+(const mint rhs) const {
    mint res(*this);
    return res += rhs;
  }
  mint operator-(const mint rhs) const {
    mint res(*this);
    return res -= rhs;
  }
  mint operator*(const mint rhs) const {
    mint res(*this);
    return res *= rhs;
  }
  friend ostream& operator<<(ostream& os,const mint rhs){
    os << rhs.x;
    return os;
  }
};

#define rep(index,num) for(int index=0;index<num;++index)

const int V = 100000;
vector<int> to[V*2];
bool visited[V*2];
vector<int> cnt;

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

int main(){
  int n;
  std::cin >> n;
  rep(i,n){
    int x,y;
    cin >> x >> y;
    y += V;
    to[x].push_back(y);
    to[y].push_back(x);
  }
  ll ans = 0;
  rep(i,V*2){
    if(visited[i]) continue;
    cnt = vector<int>(2);
    dfs(i);
    ans += (ll)cnt[0]*cnt[1];
  }
  ans -= n;
  cout << ans << endl;
  return 0;
}