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>
#define ll long long
#define ld long double
#define x first
#define y second
#define pb push_back
using namespace std;

ll n,dsu[100010],ans,tam[100010];
pair< ll , ll > p[100010];
set< ll > X[100010] , Y[100010] , labels;

ll getF( ll nd ){
    if( dsu[nd]==nd ) return nd;
    return dsu[ nd ] = getF( dsu[ nd ] );
}

void join(ll u,ll v){
    ll fu = getF( u ) , fv = getF( v );
    if( fu == fv ) return;
    dsu[ fu ] = fv;
    return;
}

int main()
{

    ios_base::sync_with_stdio(false);
    cin.tie(0);

    cin >> n;
    for(ll i=1; i<=100000; i++) dsu[ i ] = i;
    for(ll i=0; i<n; i++) cin >> p[i].x >> p[i].y , join( p[i].x , p[i].y );

    for(ll i=0; i<n; i++) X[ getF( p[i].x ) ].insert( p[i].x ) , Y[ getF( p[i].y ) ].insert( p[i].y ) , labels.insert( getF( p[i].x ) ) , tam[ getF(p[i].x) ]++;

    for( auto l : labels ) ans+=X[l].size()*Y[l].size()-tam[l];
    cout << ans;

    return 0;
}