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 Ruby (2.3.3) (judge verdict: Accepted)

N = gets.to_i
X, Y = N.times.map{gets.split.map(&:to_i)}.transpose

MX = 100001
edges = Array.new(2*MX){[]}
N.times do |i|
    edges[X[i]].push(Y[i]+MX)
    edges[Y[i]+MX].push(X[i])
end

ans = -N
visited = []
(2*MX).times do |i|
    que = [i]
    cnt = [0, 0]
    while que.size > 0
        x = que.pop
        next if visited[x]
        visited[x] = true
        cnt[x/MX] += 1
        que.concat(edges[x])
    end
    ans += cnt[0]*cnt[1]
end
puts ans