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: Accepted)

#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;

const int V = 100005;
vector<int> to[V * 2];
vector<int> cnt;
bool visited[V * 2];

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

int main(int argc, char *argv[])
{
	int N;

	cin >> N;

	for (int i = 0; i < N; i++) {
		int x, y;
		cin >> x >> y;
		y += V;
		to[x].push_back(y);
		to[y].push_back(x);
	}

	ll ans = 0;
	for (int i = 0; i < V * 2; i++) {
		if (visited[i]) {
			continue;
		}
		cnt = vector<int>(2);
		dfs(i);
		ans += (ll) cnt[0] * cnt[1];
	}
	ans -= N;
	cout << ans << endl;

	return 0;
}