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();
		
		UnionFind uf = new UnionFind(n);
		for(int i = 0; i < m; i++){
		    int x = sc.nextInt() - 1;
		    int y = sc.nextInt() - 1;
		    int z = sc.nextInt();
		    
		    uf.union(x, y);
		}
		
		// 計算
		int result = 0;
		HashSet<Integer> set = new HashSet<Integer>();
		for(int i = 0; i < n; i++){
		    set.add(uf.find(i));
		}
		result = set.size();
		
		// 出力
        System.out.println(result);
	}
}

class UnionFind {
    int[] parent;
    int[] rank;
    
    UnionFind(int n){
        this.parent = new int[n];
        this.rank = new int[n];
        for(int i = 0; i < n; i++){
            parent[i] = i;
            rank[i] = 0;
        }
    }
    
    public int find(int i){
        if(i == parent[i]) return i;
        parent[i] = find(parent[i]);
        return parent[i];
    }
    
    public void union(int x, int y){
        int xRoot = find(x);
        int yRoot = find(y);
        if(rank[xRoot] < rank[yRoot]){
            parent[xRoot] = yRoot;
        }else if(rank[xRoot] > rank[yRoot]){
            parent[yRoot] = xRoot;
        }else{
            parent[yRoot] = xRoot;
            rank[xRoot]++;
        }
    }
    
    public boolean same(int x, int y){
        return find(x) == find(y);
    }
    
}