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

ns = []
list_x = []
list_y = []
N.times{|i|
	pos = gets.strip.split(/ /).collect(&:to_i)
	ns << pos
	x, y = pos
	if l = list_x[x]
		l << i
	else
		list_x[x] = [i]
	end
	if l = list_y[y]
		l << i
	else
		list_y[y] = [i]
	end
}

groups = N.times.to_a

def get_root(groups, g)
	parent = groups[g]
	if parent == g
		return parent
	else
		return groups[g] = get_root(groups, parent)
	end
end

def set_root(groups, g, root)
	parent = groups[g]
	groups[g] = root
	if parent == g
		return
	else
		set_root(groups, parent, root)
	end
end

def combine(groups, g1, g2)
	root = get_root(groups, g1)
	set_root(groups, g2, root)
end

def combine_groups(groups, l)
	if l
		if l.length > 1
			i1 = l[0]
			(1...l.length).each{|i|
				i2 = l[i]
				combine(groups, i1, i2)
				i1 = i2
			}
		end
	end
end

list_x.each{|l|
	combine_groups(groups, l)
}

list_y.each{|l|
	combine_groups(groups, l)
}

table = {}
groups.each_with_index{|g, i|
	g = get_root(groups, g)
	if l = table[g]
		l << i
	else
		table[g] = [i]
	end
}

count = 0
table.values.each{|list|
	x_table = {}
	y_table = {}
	list.each{|i|
		x, y = ns[i]
		x_table[x] = true
		y_table[y] = true
	}
	x_count = x_table.length
	y_count = y_table.length
	if x_count >= 2 && y_count >= 2
		count += (x_count * y_count - list.length)
	end
}
puts count