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;

#ifdef DEBUG
#include "../include/debug.hpp"
#else
#define debug(...) 42
#define rdfile(...) 42
#endif

const int V = 1e5;
bool visited[V*2];

void dfs(vector<vector<int>> &to, int u, int &xn, int &yn) {
    if (visited[u]) return;
    visited[u] = true;
    if (u < V) xn ++;
    else yn ++;
    for (int v: to[u]) {
        dfs(to, v, xn, yn);
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    // rdfile("_input.txt");

    int N;
    cin >> N;
    int x, y;
    vector<vector<int>> to(V*2);
    for (int i=0; i<N; ++i) {
        cin >> x >> y;
        x--; y--;
        to[x].push_back(V+y);
        to[V+y].push_back(x);
    }

    long long ans = 0;
    for (int i=0; i<V; ++i) {
        if (visited[i]) continue;
        int xn = 0, yn = 0;
        dfs(to, i, xn, yn);
        ans += (long long)xn * yn;
    }

    cout << ans - N << endl;

    return 0;
}