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 <iostream>
#include <iomanip>
#include <vector>
#include <set>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <numeric>
#include <list>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#define rep(i, n) for (int i = 0; i < n; i++)
#define rep1(i, n) for (int i = 1; i <= n; i++)
#define show(a, n) rep(i,n) {cout<<a[i]<<' ';} cout<<endl;

using namespace std;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef long long ll;
typedef pair<long long, long long> pll;

const int INF = 1 << 30;
const long long INFL = 1LL << 62;
const int MOD = 1000000007;
const int MAX = 100000;
const int N = 100010;

vi v[2][N];
int l[2][N];
const int X = 0;
const int Y = 1;

int main() {
	int n;
	cin >> n;
	rep(i, n) {
		int x, y;
		cin >> x >> y;
		v[X][x].push_back(y);
		v[Y][y].push_back(x);
	}

	rep(i, N) {
		l[X][i] = 0;
		l[Y][i] = 0;
	}

	int label = 1;
	queue<pii> que;
	for(int x = 1; x <= MAX; x++) {
		if(v[X][x].size() == 0 || l[X][x] != 0) {
			continue;
		}
		l[X][x] = label;
		que.push(pii(x, X));
		while(!que.empty()) {
			pii p = que.front();
			que.pop();
			for(auto i : v[p.second][p.first]) {
				if(l[1 - p.second][i] == 0) {
					l[1 - p.second][i] = label;
					que.push(pii(i, 1 - p.second));
				}
			}
		}
		label++;
	}
	
	map<int, int> count_x, count_y;
	for(int i = 0; i < label; i++) {
		count_x[i] = 0;
		count_y[i] = 0;
	}
	for(int i = 1; i <= MAX; i++) {
		count_x[l[X][i]]++;
		count_y[l[Y][i]]++;
	}
	ll count_p = 0;
	for(int i = 1; i < label; i++) {
		count_p += count_x[i] * count_y[i];
	}
	cout << count_p - n << endl;
}