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++ (GCC 9.2.1) (judge verdict: Accepted)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 100010001;
const ll LINF = (ll)INF*INF*10;
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true);}
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true);}
template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second;}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << ' ' << p.second;}

struct UnionFind {
	vector<int> par; // 親を指すvector,-par[親]は木のサイズ
	UnionFind(int n):par(n, -1) {} // uniteで親を埋め込んでいく必要あり
	int root(int x) { // 親をたどる&データの整理
		if(par[x] < 0) return x;
		return par[x] = root(par[x]);
	}
	bool unite(int x, int y) { // データの結合
		x = root(x);
		y = root(y);
		if(x == y) return false;
		if(par[x] > par[y]) swap(x, y);
		par[x] += par[y];
		par[y] = x;
		return true;
	}
	bool same(int x, int y) {return root(x) == root(y);} // 所属判定
	int size(int x) {return -par[root(x)];} // 木のサイズ
};

const int N = 1e5+4;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	vi x(n), y(n);
	rep(i, n) cin >> x[i] >> y[i];
	UnionFind uft(N<<1);
	rep(i, n) uft.unite(x[i], y[i]+N);
	map<int, vvi> mp;
	map<int, int> cnt;
	rep(i, n) {
		int r = uft.root(x[i]);
		if(!mp.count(r)) {
			mp[r].assign(2, vi());
		}
		mp[r][0].emplace_back(x[i]);
		mp[r][1].emplace_back(y[i]);
		cnt[r]++;
	}

	ll ans = 0;
	for(P x:cnt) {
		vi &v0 = mp[x.first][0];
		vi &v1 = mp[x.first][1];
		sort(all(v0));
		sort(all(v1));
		v0.erase(unique(all(v0)), v0.end());
		v1.erase(unique(all(v1)), v1.end());
		ans += ll(v0.size())*ll(v1.size())-x.second;
	}

	cout << ans << endl;
}