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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
//#include <boost/multiprecision/cpp_int.hpp>
//using multiInt = boost::multiprecision::cpp_int;

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename Q_temp>
using smaller_queue = priority_queue <Q_temp, vector<Q_temp>, greater<Q_temp> >;

const int INF = (int) 1e9;
const ll LINF = (ll) 4e18;
const ll MOD = (ll) (1e9 + 7);
const double PI = acos(-1.0);

#define REP(i,m,n) for(ll i = m; i < (ll)(n); ++i)
#define rep(i,n) REP(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbg2(x_) for(auto a_ : x_) cerr << a_ << " "; cerr << endl;
#define dbg3(x_ , sx_) rep(i, sx_) cerr << x_[i] << " "; cerr << endl;
vector<int> Dx = {0, 0, -1, 1, -1, 1, -1, 1, 0};
vector<int> Dy = {1, -1, 0, 0, -1, -1, 1, 1, 0};

inline ll CEIL(ll a, ll b) {
    return (a + b - 1) / b;
}

void finish() {
    cout << -1 << endl;
    exit(0);
}

//------------------------------------------------------

class UnionFind {
private:
    vector<int> par, sizes;

public:

    UnionFind(int n = 100010) : par(n), sizes(n, 1) {
        rep(i, n) par[i] = i;
    }

    int root(int x) {
        if (x == par[x]) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) {
        x = root(x);
        y = root(y);
        if (x == y) return;
        if (sizes[x] < sizes[y]) swap(x, y);
        par[y] = x;
        sizes[x] += sizes[y];
    }

    bool same(int x, int y) {
        return root(x) == root(y);
    }

    int size(int x) {
        return sizes[root(x)];
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    UnionFind uf(n);
    vector<pair<pii, int> > v(n);

    rep(i, n) {
        cin >> v[i].first.first >> v[i].first.second;
        v[i].second = i;
    }
    sort(all(v));

    rep(i, n - 1) {
        if (v[i].first.first == v[i + 1].first.first) uf.unite(v[i].second, v[i + 1].second);
    }

    rep(i, n) swap(v[i].first.first, v[i].first.second);
    sort(all(v));

    rep(i, n - 1) {
        if (v[i].first.first == v[i + 1].first.first) uf.unite(v[i].second, v[i + 1].second);
    }

    rep(i, n) swap(v[i].first.first, v[i].first.second);

    vector<int> num[100010];
    rep(i, n) num[uf.root(v[i].second)].push_back(v[i].second);
    ll ans = 0;

    rep(i, 100010) {
        set<int> x, y;
        for (auto t : num[i]) {
            x.insert(v[t].first.first);
            y.insert(v[t].first.second);
        }
        if (x.size() == 1 || y.size() == 1) continue;
        ans += (ll) x.size() * (ll) y.size() - (ll) num[i].size();
    }
    cout << ans << endl;

    return 0;
}