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(n)
    @n = n
    @parent = Array.new(n){|i|i}
    @rank = Array.new(n, 0)
  end

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

  def unite(x, y)
    x = find(x)
    y = find(y)
    return if x == y
    if @rank[x] < @rank[y]
      @parent[x] = y
    else
      @parent[y] = x
      @rank[x] += 1 if @rank[x] == @rank[y]
    end
  end

  def same?(x, y)
    find(x) == find(y)
  end

  def num_of_cimponent
    h=Hash.new(0)
    @n.times{|i|
      h[find(i)]+=1
    }
    h.keys.size
  end
end

n,m=gets.split.map(&:to_i)
uf=UnionFind.new(n)
m.times{
  x,y,z=gets.split.map(&:to_i)
  uf.unite(x-1,y-1)
}

puts uf.num_of_cimponent