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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;


public class Main{

	public static void main(String[] args) {
		Scanner reader = new Scanner(System.in);
		int N = reader.nextInt();
		int M = reader.nextInt();
		int[] arr = new int[N+1];
		Arrays.fill(arr, -1);
		Set<Integer> rootNodes = new HashSet<Integer>();

		for (int i = 0; i < M; i++) {
			int x = reader.nextInt();
			int y = reader.nextInt();
			int z = reader.nextInt();
			if (find(arr, x) != find(arr, y))
				union(arr, x, y);
		}
		for (int i = 1; i <= N; i++) {
			rootNodes.add(find(arr, i));
		}

		System.out.print(rootNodes.size());

		reader.close();

	}

	static int find(int[] parent, int i) {
	    if (parent[i] == -1) {
	        return i;
	    }
	    return find(parent, parent[i]);
	}

	// Naive implementation of union()
	static void union(int[] parent, int x, int y) {
	    int xset = find(parent, x);
	    int yset = find(parent, y);
	    parent[xset] = yset;
	}

}