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;

namespace ABC126_E___1_or_2
{
    class Program
    {
        static void Main(string[] args)
        {
            var NM = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
            var union = new UnionFind(NM[0]);
            for (int i = 0; i < NM[1]; i++)
            {
                var xyz = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
                union.unite(xyz[0], xyz[1]);
            }
            Console.WriteLine(union.UnionCount);
        }
    }
    public class UnionFind
    {
        private int[] parents;
        // 同じ根に属する枝の数を格納する
        private int[] rootCount;
        public int UnionCount;

        public int RootCount(int branch)
        {
            return rootCount[root(branch)];
        }

        public UnionFind(int count)
        {
            parents = new int[count + 1];
            rootCount = new int[count + 1];
            UnionCount = count;
            for (int i = 0; i < count + 1; i++)
            {
                parents[i] = i;
                rootCount[i] = 1;
            }
        }

        private int root(int N)
        {
            if (parents[N] == N) return N;
            return parents[N] = root(parents[N]);
        }

        //木を併合する際に、枝の数も合算する
        public void unite(int X, int Y)
        {
            int rootX = root(X);
            int rootY = root(Y);
            if (rootX == rootY) return;
            rootCount[rootX] += rootCount[rootY];
            parents[rootY] = rootX;
            UnionCount -= 1;
        }

        public bool sameRoot(int X, int Y)
        {
            return root(X) == root(Y);
        }

    }
}