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.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[] visited = new int[n+1];
		ArrayList<Edge>[] list = new ArrayList[n+1];
		for(int i = 1; i <= n; i++){
		    list[i] = new ArrayList<Edge>();
		}
		for(int i = 0; i < m; i++){
		    int x = sc.nextInt();
		    int y = sc.nextInt();
		    int z = sc.nextInt();
		    
		    list[x].add(new Edge(x, y));
		    list[y].add(new Edge(y, x));
		}
		
		// 計算
		int result = 0;
		ArrayDeque<Integer> ad = new ArrayDeque<Integer>();
		for(int i = 1; i <= n; i++){
		    if(visited[i] == 1) continue;
		    visited[i] = 1;
		    result++;
		    ad.add(i);
		    while(!ad.isEmpty()){
		        int now = ad.poll();
		        for(Edge e : list[now]){
		            int j = e.to;
		            if(visited[j] == 1) continue;
		            visited[j] = 1;
		            ad.add(j);
		        }    
		    }
		}
		
		// 出力
        System.out.println(result);
	}
}

class Edge {
    int from;
    int to;
    
    Edge(int from, int to){
        this.from = from;
        this.to = to;
    }

    public String toString(){
        return "from:" + from + " to:" + to ;
    }
    
}