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;

long long N, X, Y, COUNT = 0;
static bool visited[200000] = {};
static vector<long long> V[200000];

void DFS(int P) {
	visited[P] = true;
	if (P < 100000) X++;
	else Y++;
	for (int R : V[P]) {
		if (!visited[R]) DFS(R);
	}
}

int main() {
	cin >> N;
	for (int i = 0; i < N; i++) {
		int x, y;
		cin >> x >> y;
		x--, y+=99999;
		V[x].push_back(y);
		V[y].push_back(x);
	}
	for (int i = 0; i < 100000; i++) {
		X = 0, Y = 0;
		if (!visited[i]) DFS(i);
		COUNT += X * Y;
	}
	cout << COUNT - N;
}