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>
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cstdio>
#include<cmath>
using namespace std;

#define ll long long
#define debug(x) cerr << "\n" \
                      << #x << ": " << x << "\n"
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define all(x) (x).begin(), (x).end()
#define int long long
#define INF 1e9
#define pii pair<int, int>
#define pt(xx) cout << "\n" \
                    << xx << "\n"
#define debugv(x, n)               \
    cout << "\n"                   \
         << #x << ": ";            \
    rep(i, n) cout << x[i] << " "; \
    cout << endl
#define debug2d(x, m, n)            \
    cout << "\n"                    \
         << #x << " is: \n";        \
    for (int i = 0; i < m; ++i)     \
    {                               \
        for (int j = 0; j < n; ++j) \
        {                           \
            cout << x[i][j] << " "; \
        }                           \
        cout << endl;               \
    }
#define nl cout << endl

#define MOD 1000000007

int N, K, L;
int ans;

const int V = 100005;
vector<vector<int>> v(2*V);
vector<int>visited(2*V,0);
int cnt[2];

void dfs(int n){
    if(visited[n]) return;
    visited[n] = 1;

    if (n < V) cnt[0]++;
    else cnt[1]++;

    for (auto x: v[n])
    {
        dfs(x);
    }
}

signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    //==============================================

    ans=0;
    cin>>N;
    int x,y;
    rep(i,N){
        cin>>x>>y;
        y = y + V;
        v[x].push_back(y);
        v[y].push_back(x);
    }

    rep(i,2*V){
        //if(v[i].size() < 1)continue;
        if(visited[i])continue;
        cnt[0] = 0;
        cnt[1] = 0;

        dfs(i);
        ans += cnt[0] * cnt[1];
    }

    ans -= N;
    cout << ans << endl;

    return 0;
}