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
import java.util.Scanner;
import java.util.ArrayList;
public class Main{
	static boolean[] visited=new boolean[100000*2];
	static ArrayList<ArrayList<Integer>> list=new ArrayList<>();
	static long xc;
	static long yc;
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		for(int i=0; i<100000*2; i++){
			ArrayList<Integer> tmp=new ArrayList<>();
			list.add(tmp);
		}
		int[] x=new int[n];
		for(int i=0; i<n; i++){
			x[i]=sc.nextInt()-1;
			int y=sc.nextInt()-1+100000;
			list.get(x[i]).add(y);
			list.get(y).add(x[i]);
		}
		long sum=0;
		for(int i:x){
			xc=0;
			yc=0;
			dfs(0,i);
			sum+=xc*yc;
		}
		System.out.println(sum-n);
	}
	static void dfs(int count,int i){
		if(!visited[i]){
			visited[i]=true;
			if(count%2==0){
				xc++;
			}else{
				yc++;
			}
			for(int j:list.get(i)){
				dfs(count+1,j);
			}
		}
	}
}
