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 <vector>
using namespace std;
const int N = 100005;
vector<int> to[2 * N];
typedef unsigned long long ull;
int cntx, cnty;
bool visited[N * 2];
void dfs(int v) {
	if (visited[v]) {
		return;
	} 
	visited[v] = true;
	if (v < N) {
		cntx++;
	}
	else {
		cnty++;
	}
	for (int u : to[v]) {
		dfs(u);
	}
	return;
}
int main()
{
	int n, i, a, b;
	cin >> n;
	for (i = 0; i < n; i++) {
		cin >> a >> b;
		b += N;
		to[a].push_back(b);
		to[b].push_back(a);
	}
	ull ans = 0;
	for (i = 1; i < 2 * N; i++) {
		if (visited[i]) continue;
		cntx = cnty = 0;
		dfs(i);
		ans += (ull)cntx * (ull)cnty;
	}
	cout << ans - n << endl;
}