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 rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}

const int MAX_N = 100100;
int n;



// n要素で初期化
void init(int par[], int rank[], int n){
    for(int i=0;i<n;i++){
        par[i] = i;
        rank[i] = 0;
    }
}

// 木の根を求める
int find(int par[], int x){
    if(par[x] == x){
        return x;
    }else{
        return par[x] = find(par,par[x]);
    }
}

// xとyの属する集合を併合
void unite(int par[], int rank[], int x, int y){
    x = find(par,x);
    y = find(par,y);
    if(x == y) return;

    if(rank[x] < rank[y]){
        par[x] = y;
    }else{
        par[y] = x;
        if(rank[x] == rank[y]) rank[x]++;
    }
}

// xとyが同じ集合に属するか否か
bool same(int par[], int x, int y){
    return find(par,x) == find(par,y);
}

int main() {
    cin >> n;
    vector<P> v; 
    vector<int> x[MAX_N];
    vector<int> y[MAX_N];

    int parx[MAX_N]; // 親
    int pary[MAX_N];
    int rankx[MAX_N]; // 木の深さ
    int ranky[MAX_N];

    init(parx,rankx,MAX_N);
    init(pary,ranky,MAX_N);

    rep(i,n){
        int x_,y_;
        cin >> x_ >> y_;
        x[x_].push_back(y_);
        y[y_].push_back(x_);
        v.push_back(P(x_,y_));
    }

    sort(v.begin(),v.end());
    ll ans = 0;

    rep(i,n){
        if(x[v[i].first].size()>=2&&y[v[i].second].size()>=2){
            //ans += (x[v[i].first].size()-1)*(y[v[i].second].size()-1);
            rep(j,x[v[i].first].size()-1){
                unite(pary,ranky,x[v[i].first][j],x[v[i].first][j+1]);
            }
            rep(j,y[v[i].second].size()-1){
                unite(parx,rankx,y[v[i].second][j],y[v[i].second][j+1]);
            }
        }
    }

    ll countx[MAX_N] = {};
    ll county[MAX_N] = {};

    rep(i,MAX_N){
        countx[find(parx,i)]++;
        county[find(pary,i)]++;
    }
    /*
    rep(i,15){
        cout << find(parx,i) << ' ';
    }
    cout << endl;
    rep(i,15){
        cout << find(pary,i) << ' ';
    }
    cout << endl;
    */


    int flag[MAX_N] = {};

    rep(i,n){
        if(countx[find(parx,v[i].first)]>=2&&county[find(pary,v[i].second)]>=2){
            ans--;
            if(flag[find(parx,v[i].first)]==0){
                flag[find(parx,v[i].first)] =1;
                ans += countx[find(parx,v[i].first)]*county[find(pary,v[i].second)];
                //cout << countx[find(parx,v[i].first)]*county[find(pary,v[i].second)] << endl;
            }
        }
    }

    cout << ans << endl;
    return 0;
}