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.

Examples

Input

3 1
1 2 1


Output

2


Input

6 5
1 2 1
2 3 2
1 3 3
4 5 4
5 6 5


Output

2


Input

100000 1
1 100000 100


Output

99999
import java.util.Scanner;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Queue;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt(), M = sc.nextInt();
		List<List<Integer>> graph = new ArrayList<>();
		for (int i = 0;i < N;++ i) graph.add(new ArrayList<>());
		for (int i = 0;i < M;++ i) {
			int x = sc.nextInt() - 1, y = sc.nextInt() - 1, z = sc.nextInt();
			graph.get(x).add(y);
			graph.get(y).add(x);
		}
		int ans = 0;
		boolean bfs[] = new boolean[N];
		Arrays.fill(bfs, false);
		for (int i = 0;i < N;++ i) {
			if (bfs[i]) continue; // 調べ終わっている
			bfs[i] = true;
			++ ans;
			Queue<Integer> queue = new ArrayDeque<>();
			for (queue.add(i);!queue.isEmpty();) {
				int tmp = queue.poll();
				for (int next : graph.get(tmp)) {
					if (bfs[next]) continue; // 既に色を塗った
					bfs[next] = true;
					queue.add(next);
				}
			}
		}
		System.out.println(ans);
	}
}