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: Accepted)

#include <bits/stdc++.h>
#define sz(c)      int(c.size())
#define rep(i,a,b) for (int i=a; i<(b); ++i)
#define per(i,a,b) for (int i=(b)-1; i>=(a); --i)
using namespace std;
using ll = long long;

struct DSU {
	vector<int> r;
	DSU(int n):r(n,-1) {}
	int get(int x) {
		return r[x]<0 ? x : r[x]=get(r[x]);
	}
	int size(int x) {
		return -r[get(x)];
	}
	bool join(int x,int y) {
		x=get(x);
		y=get(y);
		if (x==y) return false;
		if (r[x]>r[y]) swap(x,y);
		r[x]+=r[y];
		r[y]=x;
		return true;
	}
};

int const MAXN=110000;
int N;
int x[MAXN],y[MAXN];
vector<int> by_x[MAXN];
set<int> by_y[MAXN];

int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);
	cout<<fixed<<setprecision(10);

	cin>>N;
	rep(i,0,N) {
		cin>>x[i]>>y[i];
		by_x[x[i]].push_back(y[i]);
	}

	DSU ds(MAXN);
	rep(i,1,MAXN) rep(j,1,sz(by_x[i])) ds.join(by_x[i][j],by_x[i][j-1]);
	
	rep(i,0,N) by_y[ds.get(y[i])].insert(x[i]);
	
	ll res=0;
	rep(i,1,MAXN) if (ds.get(i)==i) res+=ll(ds.size(i))*sz(by_y[i]);
	cout<<res-N<<"\n";
}