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: Time Limit Exceeded)

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

set<int> x[100005],y[100005],p,q;
set<pair<int,int>> memo;
long long ans=0LL;

void dfs(int nx,int ny)
{
	for (auto i = x[nx].begin(); i != x[nx].end(); ++i)
	{
		if (memo.count(make_pair(nx,*i))==0)
		{
			q.insert(*i);
			memo.insert(make_pair(nx,*i));
			dfs(nx,*i);
		}
	}
	for (auto i = y[ny].begin(); i != y[ny].end(); ++i)
	{
		if (memo.count(make_pair(*i,ny))==0)
		{
			p.insert(*i);
			memo.insert(make_pair(*i,ny));
			dfs(*i,ny);
		}
	}
}

int main(int argc, char const *argv[])
{
	int n;
	cin>>n;
	int a[n],b[n];
	for (int i = 0; i < n; ++i)
	{
		cin>>a[i]>>b[i];
		x[a[i]].insert(b[i]);
		y[b[i]].insert(a[i]);
	}
	for (int i = 0; i < n; ++i)
	{
		if (memo.count(make_pair(a[i],b[i]))==0)
		{
			memo.insert(make_pair(a[i],b[i]));
			p.insert(a[i]);
			q.insert(b[i]);
			dfs(a[i],b[i]);
			ans+=(long long)p.size()*q.size();
			p={};q={};
		}
	}
	cout<<ans-n<<endl;
	return 0;
}