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
// author: Saman Mahdanian
#include <bits/stdc++.h>

using namespace std;
const int N = 1e5 + 10;

int n, x[N], y[N], par[N];
vector <int> xs[N], ys[N];
set <int> st[N];

int root (int);
void merge (int, int);

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> x[i] >> y[i];
		xs[x[i]].push_back(i);
		ys[y[i]].push_back(i);
	}

	iota (par, par + n, 0);
	for (int i = 1; i < N; i++) {
		for (int j: xs[i])
			merge(xs[i][0], j);
		for (int j: ys[i])
			merge(ys[i][0], j);
	}

	for (int i = 0; i < n; i++)
		st[root(i)].insert(i);

	long long ans = 0;
	for (int i = 0; i < n; i++) 
		if (st[i].size()) {
			set <int> cx, cy;
			for (int j: st[i]) {
				cx.insert(x[j]);
				cy.insert(y[j]);
			}
			ans += 1LL * cx.size() * cy.size();
		}
	cout << ans - n << endl;
}

int root (int v) {
	return par[v] == v? v : (par[v] = root(par[v]));
}

void merge (int v, int u) {
	par[root(v)] = root(u);
}

// SamMHD :: Jun27-2019 :: Another Day with Shayan...