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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
class Program
{

static string[] input = Console.ReadLine().Split(' ');
static long n = long.Parse(input[0]);
static long q = long.Parse(input[1]);
static long[] parents = new long[n];//親
static long answer = 1;
  
	static void Main()
	{
    for(long i = 1; i < n; i++)
    {
    parents[i] = i;
    }
    
    for(long i = 0; i < q; i++)
    {
    long[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);
        if(whatParent(nums[0]-1) != whatParent(nums[1]-1))
          parents[whatParent(nums[0]-1)] = whatParent(nums[1]-1);//親を結合して統一。親が2人できることはない。
    }

    for(long i = 0; i < n-1; i++)
    {
      if(whatParent(i) != whatParent(i+1))//別の島
      {
        answer++;
        parents[whatParent(i)] = whatParent(i+1);
      }
    }

    Console.WriteLine(answer);
	}

  static long whatParent(long child)
  {
    if(parents[child] == child) return child;
    else return parents[child] = whatParent(parents[child]);
  }
  
}