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.Collections.Generic;
using System.Text;
using System.Numerics;

namespace SolutionCS
{
    class Program
    {
        class IntUnionFind
        {
            // 親の要素を指す配列
            int[] rootArray;

            // 初期化(最大要素数指定する)
            public IntUnionFind(int max_element)
            {
                rootArray = new int[max_element];
                for (int i = 0; i < rootArray.Length; i++)
                {
                    rootArray[i] = i;       // 自分自身を指す=root
                }
            }

            public bool IsRoot(int element)
            {
                return rootArray[element] == element;
            }

            // 要素の根を返す
            public int Root(int element)
            {
                // 要素インデックス=親の要素インデックスなら親、そうでないなら親をたどる
                if (rootArray[element] != element)
                {
                    // 根を取得
                    int r = Root(rootArray[element]);
                    // 根を付け替え
                    rootArray[element] = r;
                    // 根を返す
                    return r;
                }

                // 根を返す
                return element;
            }

            // 同じ根かどうかを返す
            public bool IsSameRoot(int element1, int element2)
            {
                return (Root(element1) == Root(element2));
            }

            // 要素を結合
            public void Unite(int element1, int element2)
            {
                int root1 = Root(element1);
                int root2 = Root(element2);

                if (root1 == root2)
                {
                    return;
                }

                rootArray[root1] = root2;
            }
        }


        static void Main(string[] args)
        {
            string[] inS = Console.ReadLine().Split(' ');
            int N = int.Parse(inS[0]);
            int M = int.Parse(inS[1]);

            IntUnionFind uf = new IntUnionFind(N);

            for (int i = 0; i < M; i++)
            {
                string[] inP = Console.ReadLine().Split(' ');
                int e0 = int.Parse(inP[0]) - 1;
                int e1 = int.Parse(inP[1]) - 1;

                uf.Unite(e0, e1);
            }

            int result = 0;

            for (int i = 0; i < N; i++)
            {
                if (uf.IsRoot(i))
                {
                    result++;
                }
            }

            Console.WriteLine("{0}", result);
        }
    }
}