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;

class UnionFindTree {
	int[] par;
	int[] rank;
	int[] peers; //仲間の数
	int[] dis; //根からの距離

	public UnionFindTree(int n){
		par = new int[n];
		rank = new int[n];
		peers = new int[n];

		for(int i=0;i<n;i++){
			par[i] = i;
			peers[i] = 1;
		}
	}

	//木の根を求める
	int find(int x){
		if(par[x] == x){
			return x;
		}
		else{
			return par[x] = find(par[x]);
		}
	}

	//xとyの属する集合を併合
	void unite(int x, int y){
		int px = find(x);
		int py = find(y);

		if(px == py){
			return;
		}
		else if(rank[px] < rank[py]){
			peers[py] += peers[px];
			par[px] = py;

		}
		else{
			peers[px] += peers[py];
			par[py] = px;
			if(rank[px]==rank[py]){
				rank[px]++;
			}
		}

	}


	//xとyが同じ集合に属するか
	boolean isEquivalent(int x, int y){
		return find(x) == find(y);
	}

	//xの仲間の数を求める
	int peersnum(int x){
		return peers[find(x)];
	}
}

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();
		int M = sc.nextInt();

		UnionFindTree uf = new UnionFindTree(N);

		for(int i = 0 ; i < M ; i++){
			int X = sc.nextInt()-1;
			int Y = sc.nextInt()-1;
			int Z = sc.nextInt();

			uf.unite(X,Y);
		}
		int count = 0;
		for(int i = 0 ; i < N ; i++){
			if(uf.find(i) == i){
				count++;
			}
		}
		System.out.println(count);
		sc.close();
	}

}