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.*;

public class Main {
	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	ArrayList<ArrayList<Integer>> g = new ArrayList<>();
	boolean used[];
	public void run() {
		Scanner sc = new Scanner(System.in);
		int N=sc.nextInt();
		int M=sc.nextInt();
		for(int i=0; i<N; i++) {
			g.add(new ArrayList<Integer>());
		}
		used = new boolean[N];
		for(int i=0; i<M; i++) {
			int x=sc.nextInt(); x--;
			int y=sc.nextInt(); y--;
			int z=sc.nextInt();
			g.get(x).add(y);
			g.get(y).add(x);
		}
		int cnt=0;
		for(int i=0; i<N; i++) {
			if(!used[i]) {
				dfs(i);
				cnt++;
			}
		}
		System.out.println(cnt);
		sc.close();
	}

	public void dfs(int r) {
		ArrayDeque<Integer> s = new ArrayDeque<Integer>();
		s.add(r);
		used[r]=true;
		while(!s.isEmpty()) {
			int u = s.pop();
			for(int i=0; i<g.get(u).size(); i++) {
				int v = g.get(u).get(i);
				if(used[v]) continue;
				s.push(v);
				used[v]=true;
			}
		}
	}
}
