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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int f[200008],resx[200008],resy[200008];
int Mana(int x)
{
    if(x==f[x])return x;
    return f[x]=Mana(f[x]);
}
int main()
{
    long long sum=0;
    int u,v,n;
    scanf("%d",&n);
    for(int i=0;i<=200001;i++)f[i]=i;
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&u,&v);
        f[Mana(u)]=Mana(v+100000);
    }
    for(int i=1;i<=100000;i++)resx[Mana(i)]+=1;
    for(int i=100001;i<=200000;i++)resy[Mana(i)]+=1;
    for(int i=1;i<=200000;i++)sum+=(long long)resx[i]*(long long)resy[i];
    printf("%lld\n",sum-n);
}