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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define PI 3.141592653589793
#define rep(i, n) for (int i = 0; i < (n); i++)
#define REP(i, a, n) for (int i = a; i < (n); i++)
#define rrep(i, n, k) for (int i = (n); i >= (k); i--);
#define all(x) (x).begin(), (x).end()
#define vi vector<int>
template<class T> istream &operator>>(istream&is,vector<T>&v){for(auto &elemnt:v)is>>elemnt;return is;}
template<class T,class U> istream &operator>>(istream&is,pair<T,U>&p){is>>p.first>>p.second;return is;}
template<class T>vector<T> make_vector(size_t a){return vector<T>(a);}
template<class T, class... Ts>auto make_vector(size_t a, Ts... ts){return vector<decltype(make_vector<T>(ts...))>(a, make_vector<T>(ts...));}
const int MOD = 1e9 + 7;
const int INF = 2e18;
typedef bitset<100500> bt;

signed main() {
    int N;
    cin >> N;
    map<int, int> x;
    vector<pair<int, int>> input(N);
    for (int i = 0; i < N; i++) {
        cin >> input[i].first >> input[i].second;
        x[input[i].first]++;
    }
    set<int> search, count;
    for (int i = 0; i < N; i++) if (x[input[i].first] >= 2) search.insert(input[i].second);
    int minus = N;
    for (int i = 0; i < N; i++) {
        if (search.find(input[i].second) != search.end()) count.insert(input[i].first);
        else minus--;
    }
    cout << search.size() * count.size() - minus << endl;
}