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.*;

class UF{
	public int[] a;
	public UF(int size){
		Arrays.fill(a=new int[size],-1);
	}
	public int root(int t){
		return a[t]<0?t:(a[t]=root(a[t]));
	}
	public boolean connect(int p,int c){
		p=root(p);
		c=root(c);
		if(p==c)
			return false;
		if(a[p]>a[c]){
			int buf=c;
			c=p;
			p=buf;
		}
		a[p]+=a[c];
		a[c]=p;
		return true;
	}
	public boolean isConnected(int a,int b){
		return root(a)==root(b);
	}
	public int groupSize(int t){
		return -a[root(t)];
	}
}

class Main{

	public static void main(String[] $){
		Scanner s=new Scanner(System.in);
		int n=s.nextInt(),m=s.nextInt();
		UF uf=new UF(n);
		for(int i=0;i<m;++i) {
			uf.connect(s.nextInt()-1,s.nextInt()-1);
			s.nextInt();
		}
		System.out.println(Arrays.stream(uf.a).filter(i->i<0).count());
	}
}