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
  attr_accessor :parent, :rank
  
  def initialize(n)
    @parent = []
    @rank = []
    n.times do |i|
      @parent[i] = i
      @rank[i] = 0
    end
  end

  def find(x)
    if @parent[x] == x
      return x
    else
      return @parent[x] = find(@parent[x])
    end
  end

  def unite(x,y)
    x,y = find(x),find(y)
    return if x==y

    x,y = y,x if @rank[x] < @rank[y]
    @rank[x] += 1 if @rank[x] == @rank[y]
    @parent[y] = x
  end
end

n,m = gets.split.map(&:to_i)
tree = UnionFind.new(n+1)

m.times do
  x,y,z = gets.split.map(&:to_i)
  tree.unite(x,y)
end

parents = [0]*(n+1)
1.upto(n) do |i|
  parents[tree.find(i)] += 1
end
ans = 0
parents.each do |v|
  ans += 1 if v > 0
end
p ans