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
#define REP(i, n) for(int i = 0; i < (int)(n); i++)

#include "bits/stdc++.h"
using namespace std;

const int V = 100001;
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;
	cin >> N;
	REP(i, N) {
		int x, y;
		cin >> x >> y;
		y += V;
		to[x].push_back(y);
		to[y].push_back(x);
	}
	long long int ans = 0;
	REP(i, V*2) {
		if (visited[i]) continue;
		cnt = vector<int>(2);
		dfs(i);
		ans += (long long int)cnt[0] * cnt[1];
	}
	ans -= N;
	cout << ans << endl;

	return 0;
}
