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)

class Node
  attr_accessor :self, :parent
  def initialize(i)
    @self, @parent = i, i
  end
end

class UnionFindTree
  attr_accessor :nodes

  #
  def initialize(n)
    @nodes = (1..n).map{|i| Node.new(i) }
  end

  # 1番上の親を返す.
  def root(x)
    return x if x == @nodes[x-1].parent
    return @nodes[x-1].parent = root(@nodes[x-1].parent)
  end

  # 判定する
  #   2つの木が同じ親に属していればtrue
  def check(a, b)
    return true if root(a) == root(b)
    return false
  end

  # まとめる
  def union(a, b)
    return if check(a, b)
    #$ans -= 1
    @nodes[root(b)-1].parent = root(a)
  end

  #
  def parent_size
    parents = []
    @nodes.each do | node |
      parents << node.parent
    end
    return parents.uniq.size
  end
end

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

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