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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;


public class Main {

	public static void main(String[] args) {
		try {
			BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
			String[] input1 = bf.readLine().split(" ");
			int n = Integer.parseInt(input1[0]);
			int m = Integer.parseInt(input1[1]);

			int magic = 0;
			Map<Integer, Set<Integer>> convination = new HashMap<Integer, Set<Integer>>();
			boolean[] cards = new boolean[n + 1];
			for (int i = 1; i <= m; i++) {
				String[] input2 = bf.readLine().split(" ");
				int ax = Integer.parseInt(input2[0]);
				int ay = Integer.parseInt(input2[1]);
				int z = Integer.parseInt(input2[1]);
				if (!convination.containsKey(ax)) {
					convination.put(ax, new HashSet<Integer>());
				}
				if (!convination.containsKey(ay)) {
					convination.put(ay, new HashSet<Integer>());
				}
				convination.get(ax).add(ay);
			}

			// 計算式で結ばれているものは、Ax、Ayいずれか一つが分かればもう片方も分かる
			for (Map.Entry<Integer, Set<Integer>> conv : convination.entrySet()) {
				int ones = conv.getKey();
				if (!cards[ones]) {
					magic++;
					cards[ones] = true;
				}
				for (Integer other : conv.getValue()) {
					cards[other] = true;
				}
			}

			// それでも分からないカードはひとつずつ魔法を使う
			for (int i = 1; i < n + 1; i++) {
				if (!cards[i]) {
					magic++;
				}
			}
			System.out.println(magic);
		} catch (IOException e1) {
		}
	}

}