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: Accepted)

class UnionFind
  def initialize(size)
    @rank = Array.new(size, 0)
    @parent = Array.new(size, &:itself)
  end

  def unite(id_x, id_y)
    x_parent = get_parent(id_x)
    y_parent = get_parent(id_y)
    return if x_parent == y_parent

    if @rank[x_parent] > @rank[y_parent]
      @parent[y_parent] = x_parent
    else
      @parent[x_parent] = y_parent
      @rank[y_parent] += 1 if @rank[x_parent] == @rank[y_parent]
    end
  end

  def get_parent(id_x)
    @parent[id_x] == id_x ? id_x : (@parent[id_x] = get_parent(@parent[id_x]))
  end

  def same_parent?(id_x, id_y)
    get_parent(id_x) == get_parent(id_y)
  end
end

N,M=gets.split.map &:to_i
as=M.times.map{gets.split.map(&:to_i)}
tree=UnionFind.new(N)
for i in 0...M
    tree.unite(as[i][0]-1,as[i][1]-1)
end
puts (0...N).to_a.map{|x|tree.get_parent(x)}.uniq.size