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 <bits/stdc++.h>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x)+modulo)%modulo)

struct unionfind{
	vector<int> data;
	vector<int> size;
	unionfind(int n){
		for(int i=0;i<n;i++){
			data.push_back(i);
			size.push_back(1);
		}
	}
	
	int find(int x){
		if(data[x]==x)return x;
		return data[x]=find(data[x]);
	}
	
	bool unite(int x,int y){
		x=find(x);y=find(y);
		if(x==y)return false;
		data[x]=y;
		size[y]+=size[x];
		return true;
	}
	
	bool check(int x,int y){
		return (find(x)==find(y));
	}
	
	int get_size(int x){
		int X = find(x);
		return size[X];
	}
};


int main(){
	int N;
	cin>>N;
	vector<vector<int>> x(10010,vector<int>(0)),y(10010,vector<int>(0));
	vector<pair<int,int>> V(N);
	
	unionfind uf(N);
	
	for(int i=0;i<N;i++){
		int a,b;
		cin>>V[i].first>>V[i].second;
		V[i].first--;V[i].second--;
		
		if(x[V[i].first].size()!=0){
			uf.unite(i,x[V[i].first][0]);
		}
		if(y[V[i].second].size()!=0){
			uf.unite(i,y[V[i].second][0]);
		}
		
		x[V[i].first].push_back(i);
		y[V[i].second].push_back(i);
	}
	
	vector<vector<int>> P(10010,vector<int>(0));
	
	for(int i=0;i<N;i++){
		P[uf.find(i)].push_back(i);
	}
	
	long long ans = 0;
	
	for(int i=0;i<N;i++){
		set<int> xx,yy;
		for(int j=0;j<P[i].size();j++){
			xx.insert(V[P[i][j]].first);
			yy.insert(V[P[i][j]].second);
		}
		long long a = (long long)xx.size()*yy.size();
		if(a==0)continue;
		ans = (long long)ans + a - P[i].size();
	}
	cout<<ans<<endl;
	
	return 0;
}