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.Linq;
using static System.Math;
using static System.Console;

namespace E_ABC126
{
    class MainClass
    {

        static int[] s;


        public static void Main(string[] args)
        {
            s = Array.ConvertAll(ReadLine().Split().ToArray(), int.Parse);
            UnionFind u = new UnionFind(s[0]);
            for(int i=0;i<s[1];i++){
                int[] ab = new int[3];
                ab = Array.ConvertAll(ReadLine().Split().ToArray(), int.Parse);
                u.Connect(ab[0]-1,ab[1]-1);
            }
            int ans = 0;
            for(int i=0;i<s[0];i++){
               // WriteLine(u.Root(i));
                if(u.Root(i)==i){
                    ans++;
                }
            }
            WriteLine(ans);
        }
    }


     class UnionFind
    {
        public List<int> Parent;
        public UnionFind(int n)
        {
            //親の番号をそれぞれ格納する。自分が親なら、-(サイズ)
            this.Parent = Enumerable.Repeat(-1, n).ToList();
        }

        public int Size(int a)
        {
            if (Parent[a] < 0)return -Parent[a];
            else return -Parent[Root(a)];
        }

        public int Root(int a)
        {
            if (Parent[a] < 0) return a;
            else
            {
                int root = Root(Parent[a]);
                Parent[a] = root;
                return root;
            }
        }

        public bool Connect(int a, int b)
        {
            a = Root(a);
            b = Root(b);
            if (a == b) return false;
            else
            {
                //a > bにする
                if (a < b)
                {
                    int tmp = a;
                    a = b;
                    b = tmp;
                }
                //bの親をaにして、aのsizeにbのぶんを足す
                Parent[a] += Parent[b];
                Parent[b] = a;
                return true;
            }
        }
    }
}