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 <bits/stdc++.h>
using namespace std;

struct UF {
	UF *p;
	int s, a;
	UF() :p(nullptr), s(1), a(0) {}
	UF *Find() {
		if (p == nullptr)
			return this;
		return p = p->Find();
	}
	int Union(UF &n) {
		auto *r = Find(), *nr = n.Find();
		if (r == nr) return 0;
		if (r->s < nr->s) swap(r, nr);
		nr->p = r, r->s += nr->s;
		r->a += nr->a;
		return 1;
	}
} u[222222];
int main() {
	int n;
	cin >> n;
	for (int i = 0; i < 100000; i++)
		u[i + 1].a = 1;
	for (int i = 0; i < n; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		u[x].Union(u[y + 100000]);
	}
	long r = 0;
	for (int i = 0; i < 200000; i++)
		if (u[i + 1].p == nullptr)
			r += u[i + 1].a*long(u[i + 1].s - u[i + 1].a);
	cout << r - n << endl;
	return 0;
}
