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)

# 写経元
# https://atcoder.jp/contests/abc126/submissions/5487745

class Node
  attr_accessor :parent
  def initialize(n)
    @parent = n
  end
end

class UnionFindTree
  attr_accessor :nodes#, :parents

  # 木の本数分, 自分自身を親として子を作成する.
  def initialize(n)
    @nodes = (1..n).to_a.map{|i| Node.new(i)}
  end

  # x に対応する1番上の親を返す.
  #   x-1 しているのは 1..n と index を揃えるため.
  def find(x)
    while @nodes[x-1].parent != x # x と xの親 が一致すれば1番上の親.
      x = @nodes[x-1].parent      # x と xの親 が一致しなければ, xを1つ上の親に置き換える.
    end
    #return  x = @nodes[x-1].parent
    return x
    #return x if x == @nodes[x-1].parent
    #return @nodes[x-1].parent = find(@nodes[x-1].parent)
  end

  # 親が一致しない場合, bの親をaの1番上の親に置き換える.
  def union(a, b)
    return if find(a) == find(b) # 1番上の親が一致すれば終了.
    $ans -= 1
    #return @nodes[find(b)-1].parent = find(a) # find(b)-1 ????
    return @nodes[b-1].parent = find(a) # bの親をaの1番上の親に置き換えて終了
  end

  #
  #def parents
  #  @nodes.map(&:parent)
  #end

end

n, m = gets.rstrip.split.map(&:to_i)
tree = UnionFindTree.new(n)

$ans = n
m.times do
  x, y, _ = gets.split.map(&:to_i)
  tree.union(x, y)
end

p $ans