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

namespace debug
{
    class ProgramE
    {
        static void Main(string[] args)
        {

            //入力
            string[] t = Console.ReadLine().Split(' ');
            int n = int.Parse(t[0]);
            int k = int.Parse(t[1]);

            int[] ans = new int[n];
            int count = 0;
           

            UnionFind tree = new UnionFind();//UnionFindクラスを設定
            tree.par = new int[n];
            tree.temp = new int[n];
            tree.reset(n);


            for (int i = 0;i < k;i++)
            {
                string[] s = Console.ReadLine().Split(' ');
                int u = int.Parse(s[0]);
                int v = int.Parse(s[1]);
                int w = int.Parse(s[2]);
                tree.unite(u - 1, v - 1);
            }

            for (int i = 0; i < n; i++)
                ans[tree.size(i) -1]++;

            for (int i = 0; i < n; i++)
                count += ans[i] /( i+ 1);


            Console.WriteLine(count);
            return;

        }
    }

    /*UnionFind法のクラス*/
    class UnionFind
    {
 
        public int[] par;
        public int[] temp;


        //マイナスは根を示す。
        public void reset(int n)
        {
            for(int i = 0;i < n;i++)
                par[i] = -1;
        }

        //マイナスならそれはまだ属していないので根である。そうでないならさかのぼる
        public int root(int x)
        {
            if(par[x] < 0)
                return x;
            else
                return root(par[x]);
        }

        //根の併合を行い、根には要素数を足す（マイナスの値）
        public void unite(int x, int y)
        { // xとyの木を併合
            int rx = root(x); //xの根をrx
            int ry = root(y); //yの根をry
            if (rx == ry) 
                 return; //xとyの根が同じ(=同じ木にある)時はそのまま
            if(par[ry] > par[rx])
            {
                 int temp = ry;
                 ry = rx;
                 rx = temp;
            }
            par[ry] += par[rx];
            par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
        }

        // 2つのデータx, yが属する木が同じならtrueを返す
        public bool same(int x, int y) 
        { 
            int rx = root(x);
            int ry = root(y);
            return rx == ry;
        }

        //根の元に入っている
        public int size(int x)
        {
            return -par[root(x)];
        }
    }
}