Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

struct Point{
	int x;
	int y;
};

bool operator==(const Point& left, const Point& right){
	return 
		left.x == right.x && 
		left.y == right.y;
}

bool operator<(const Point& left, const Point& right){
	return 
		left.x < right.x ||
		left.y < right.y;
}

int main(){
	int n,i,j,k;
	cin >> n;
	vector<Point> p(n);
	for(i=0;i<n;i++){
		cin >> p[i].x >> p[i].y;
	}

	sort(p.begin(),p.end());
	int a=0,b,c,d,count=0;
	Point ab,cd;
	for(i=0;i<n;i++){
		vector<Point> ad,cb;
		a=p[i].x;
		b=p[i].y;
		ab=p[i];
		for(j=0;j<n;j++){
			if(j==i) continue;
			if(p[j].x==a){
				d=p[j].y;
				ad.push_back(p[j]);
			}
			if(p[j].y==b){
				c=p[j].x;
				cb.push_back(p[j]);
			}
		}
		for(Point ady : ad){
			cd.y=ady.y;
			for(Point cbx : cb){
				cd.x=cbx.x;
				if(find(p.begin(),p.end(),cd)==p.end()){
					count++;
				}
			}
		}
	}

	cout << count << endl;
	return 0;
}