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;
typedef long long ll;
#define rep(i, a, b) for(int i = a; i < b; i++)


int N;
ll cnt, cntx, cnty;
vector<bool> used(2e5, 0);
vector<vector<int> > G(1e5 * 2 + 10);

void dfs(int idx, int par){
    used[idx] = true;
    if(idx < 1e5) cntx++;
    else cnty++;
    for(auto& to: G[idx]){
        if(to == par) continue;
        cnt++;
        if(used[to]) continue;
        dfs(to, idx);
    }
}

int main(){
    cin >> N;
    rep(i, 0, N){
        int x, y; cin >> x >> y; x--, y--;
        G[x].push_back(y + 1e5);
        G[y + 1e5].push_back(x);
    }
    ll ans = 0;
    rep(i, 0, 1e5) if(!used[i]){
        cnt = cnty = cntx = 0;
        dfs(i, -1);
        if(cntx > 1 && cnty > 1){
            ans += cntx * cnty - cnt;
        }
    }
    cout << ans << endl;
}