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)

# 上位ノードほど @rank が大きい数字となる
class Node
  attr_accessor :parent, :rank

  # 親は自分自身、ランクは 0 として初期化する
  def initialize(n)
    @parent = n
    @rank = 0
  end
end

class UnionFind
  # 0 から n までのノード（つまりn + 1個）を初期化する
  def initialize(n)
    @nodes = (0..n).map { |i| Node.new(i) }
  end

  # 再帰で最も上位の親を探して更新し、その値を返す
  def parent(x)
    return x if @nodes[x].parent == x
    return @nodes[x].parent = parent(@nodes[x].parent)
  end

  # a と b を仲間にする
  def unite(a, b)
    a_parent = parent(a)
    b_parent = parent(b)
    # a 及び b の親が同じ場合は何もしない
    return if a_parent == b_parent

    # a親ランク < b親ランク の場合は、b親をa親の親とする
    # なお、ランクをいじる必要はなし
    if @nodes[a_parent].rank < @nodes[b_parent].rank
      @nodes[a_parent].parent = b_parent
    # その他の場合は、a親をb親の親とする
    else
      @nodes[b_parent].parent = a_parent
    # a親、及びb親のランクが同じ場合は、a親のランクを +1 する
      @nodes[a_parent].rank += 1 if @nodes[a_parent].rank == @nodes[b_parent].rank
    end
  end

  # a と b の親が同じかどうかを返す
  def same?(a, b)
    parent(a) == parent(b)
  end

  # 各ノードの親を全て出力する確認用
  def parents
    @nodes.map(&:parent)
  end
end

n, m = gets.split.map(&:to_i)
edges = []
m.times { edges << gets.split.map(&:to_i) }
sum = 0

tree = UnionFind.new(n)
edges.each do |edge|
  tree.unite(edge[0], edge[1])
end
p tree.parents.uniq.size - 1