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 <cstdio>
#include <cassert>
#include <algorithm>    
#include <iostream>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <cstring>
#include <cmath>
#include <bitset>
#define pb push_back
#define mk make_pair
#define int long long
#define eb emplace_back
#define eps 1e-8
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()


using namespace std;
typedef long double ld;
typedef unsigned int ui;
typedef pair<int,int> pii;
typedef tuple<int, int, int> tiii;
typedef vector<int> vii;
typedef vector<long double> vd;
const int inf = 1e9;
const int INF = 1e18;
const int M = 1e9 + 7;

const int maxn = 2e5;
pii a[maxn];
set<int> s[maxn];
std::vector<int> v[maxn], vv[maxn];
int r[maxn], p[maxn];

void init(int n){
	for(int i = 1; i <= n; ++i){
		r[i] = 0;
		p[i] = i;
	}
}

int find(int x){ 
	return x == p[x] ? x : p[x] = find(p[x]);
}

inline int same(int x, int y){ 
	return find(x) == find(y); 
}

void unite(int x, int y){
	int px = find(x), py = find(y);	
	if(px == py) {
		return;
	}
	if(r[px] < r[py]) {
		p[px] = py;
	}
	else {
		p[py] = px;
		if(r[px] == r[py]) r[px]++;
	}
}


signed main()
{
#ifndef ONLINE_JUDGE
	// freopen("in.txt", "r", stdin);
	// freopen("out.txt", "w", stdout);
#endif
	int n;
	cin >> n;
	for(int i = 0; i < n; ++i){
		cin >> a[i].fi >> a[i].se;
	}
	sort(a, a + n);
	int cur = a[0].fi, i = 0, res = 0, dep = 0;
	while(i < n){
		++dep;
		while(i < n && a[i].fi == cur){
			v[a[i].se].pb(dep);
			s[dep].insert(a[i].se);
			++i;
		}
		cur = a[i].fi;
	}
	init(1e5 + 5);
	for(int i = 1; i <= 1e5; ++i){
		if((int)v[i].size() < 2) continue;
		for(int j = 1; j < (int)v[i].size(); ++j){
			unite(v[i][j], v[i][0]);
		}
	}
	for(int i = 1; i <= 1e5; ++i){
		vv[find(i)].pb(i);
	}
	for(int i = 1; i <= 1e5; ++i){
		if((int)vv[i].size() > 1){
			set<int> t;
			int tot = 0;
			for(int j : vv[i]){
				for(int k : s[j]){
					t.insert(k);
					++tot;
				}
			}
			res += (int)vv[i].size() * (int)t.size() - tot;
		}
	}
	cout << res << endl;



    
	return 0;
}