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>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<ll,ll>;
using vl = vector<ll>;
using Map = map<ll,ll>;
using T = tuple<ll,ll,ll>;
using vvl = vector<vector<ll>>;
#define all(v) v.begin(), v.end()
#define prt(v) cout<<v<<"\n";
#define fi(v) get<0>(v)
#define se(v) get<1>(v)
#define th(v) get<2>(v)
#define endl "\n"
template <typename T> bool chmax(T &a, const T &b){if (a<b){a=b;return 1;}return 0;}
template <typename T> bool chmin(T &a, const T &b){if (a>b){a=b;return 1;}return 0;}
const ll INF=1LL<<60;
const ll MOD=1000000007;

#define MAX 100003

class UnionFind {
public:
    vector < ll > par; // 各元の親を表す配列
    vector < ll > siz; // 素集合のサイズを表す配列(1 で初期化)

    // Constructor,[0,sz_-1]で定義
    UnionFind(ll sz_): par(sz_), siz(sz_, 1) {
        for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
    }
    void init(ll sz_) {
        par.resize(sz_);
        siz.assign(sz_, 1);
        for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
    }

    // Member Function
    // Find
    ll root(ll x) { // 根の検索
        while (par[x] != x) {
            x = par[x] = par[par[x]]; // x の親の親を x の親とする
        }
        return x;
    }

    // Union(Unite, Merge)
    bool unite(ll x, ll y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;
        // merge technique（データ構造をマージするテク．小を大にくっつける）
        if (siz[x] < siz[y]) swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }

    bool same(ll x, ll y) { // 連結判定
        return root(x) == root(y);
    }

    ll size(ll x) { // 素集合のサイズ
        return siz[root(x)];
    }
};

signed main(void){
    cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);
    ll N;
    cin >> N;
    vector<ll> X(N,0);
    vector<ll> Y(N,0);
    UnionFind tree(2*MAX);
    for(ll i=0;i<N;++i){cin>>X[i]>>Y[i];tree.unite(X[i],Y[i]+MAX);}
    Map mx,my;
    for(ll i=0;i<MAX;++i)mx[tree.root(i)]++;
    for(ll i=MAX;i<2*MAX;++i)my[tree.root(i)]++;
    ll rec=0;
    for(ll i=0;i<2*MAX;++i)rec+=mx[i]*my[i];
    prt(rec-N)


    return 0;
}