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.

Examples

Input

3
1 1
5 1
5 5


Output

1


Input

2
10 10
20 20


Output

0


Input

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5


Output

16
#include<iostream>
#include<algorithm>
#include<vector>
#include<utility>
using namespace std;
using ll = long long int;
constexpr int MAX = 100005;

int n;
vector<vector<int>> to(MAX * 2);
vector<bool> visited(MAX * 2, false);
vector<ll> cnt(2, 0);
ll ret = 0;

void dfs(int v) {
	if (visited[v])
		return;
	visited[v] = true;
	cnt[v / MAX]++;
	for (int i = 0; i < to[v].size(); i++)
		dfs(to[v][i]);
	return;
}

int main(void) {
	cin >> n;
	for (size_t i = 0; i < n; i++) {
		int ai, bi;
		cin >> ai >> bi;
		bi += MAX;
		to[ai].push_back(bi);
		to[bi].push_back(ai);
	}
	ll ret = 0;
	for (int i = 0; i < MAX*2; i++) {
		if (visited[i])
			continue;
		cnt = vector<ll>(2, 0);
		dfs(i);
		ret += ll(cnt[0] * cnt[1]);
	}
	ret -= n;
	cout << ret << endl;
	return 0;
}