Problem:
Score : 500 points

Problem Statement

There are N cards placed face down in a row. On each card, an integer 1 or 2 is written.

Let A_i be the integer written on the i-th card.

Your objective is to guess A_1, A_2, ..., A_N correctly.

You know the following facts:

For each i = 1, 2, ..., M, the value A_{X_i} + A_{Y_i} + Z_i is an even number.

You are a magician and can use the following magic any number of times:

Magic: Choose one card and know the integer A_i written on it. The cost of using this magic is 1.

What is the minimum cost required to determine all of A_1, A_2, ..., A_N?

It is guaranteed that there is no contradiction in given input.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq X_i < Y_i \leq N

1 \leq Z_i \leq 100

The pairs (X_i, Y_i) are distinct.

There is no contradiction in input. (That is, there exist integers A_1, A_2, ..., A_N that satisfy the conditions.)

Input

Input is given from Standard Input in the following format:

N M
X_1 Y_1 Z_1
X_2 Y_2 Z_2
\vdots
X_M Y_M Z_M

Output

Print the minimum total cost required to determine all of A_1, A_2, ..., A_N.

Sample Input 1

3 1
1 2 1

Sample Output 1

2

You can determine all of A_1, A_2, A_3 by using the magic for the first and third cards.

Sample Input 2

6 5
1 2 1
2 3 2
1 3 3
4 5 4
5 6 5

Sample Output 2

2

Sample Input 3

100000 1
1 100000 100

Sample Output 3

99999

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

N, M = gets.strip.split(/ /).collect(&:to_i)

groups = {}
group_count = 0
appeared = []

group_id = 0

M.times{
	x, y, z = gets.strip.split(/ /).collect(&:to_i)
	
	x_id = groups[x]
	y_id = groups[y]
	if x_id == nil && y_id == nil
		groups[x] = groups[y] = group_id
		group_id += 1
		group_count += 1
	elsif x_id == nil
		y_id << x
		groups[x] = y_id
	elsif y_id == nil
		x_id << y
		groups[y] = x_id
	elsif x_id == y_id
	else
		groups[x] = groups[y] = x_id
		group_count -= 1
	end
	appeared[x] = true
	appeared[y] = true
}

puts N - appeared.count{|flag| flag == true} + group_count