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: Time Limit Exceeded)

class UnionFind
  def initialize(n)
    @par = n.times.to_a
  end

  def root(x)
    return x if @par[x] == x
    root(@par[x])
  end

  def unite(x, y)
    rx = root(x)
    ry = root(y)
    return if rx == ry
    @par[rx] = ry 
  end

  def count_tree()
    @par.map.with_index { |e, i| e == i ? 1 : 0}.reduce(&:+)
  end
end

N, M = gets.split(' ').map(&:to_i)
uf = UnionFind.new(N)

M.times do
  x, y, z = gets.split(' ').map{ |n| n.to_i - 1 }
  uf.unite(x, y)
end
puts uf.count_tree()