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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ABC126E
{
    class Program
    {
        static void Main(string[] args)
        {
            var nums = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToList();
            var card = Enumerable.Repeat(-1, nums[0]).ToList();

            foreach (var i in Enumerable.Range(0, nums[1]))
            {
                var info = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToList();
                var x = info[0] - 1;
                var y = info[1] - 1;

                // どちらかがペア登録されていれば、同じグループとして登録する
                var pairId = Math.Max(card[x], card[y]);

                // どちらもまだペア登録されていないなら、新しいグループ番号を割り当てる
                if (pairId < 0)
                {
                    pairId = i;
                }

                // いずれかが分かればOKなペアとして記録する
                card[x] = pairId;
                card[y] = pairId;
            }

            var sum = 0;
            sum += card.Count(x => x < 0);
            sum += card.Distinct().Count(x => x >= 0);

            Console.WriteLine(sum);
        }
    }
}