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 java.io.*;
 
public class Main {
    public static void main(String[] args) throws Exception {
        FastScanner sc = new FastScanner(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()%2;
            uf.union(x,y);
        }
        HashSet<Integer> hs = new HashSet<Integer>();
        for(int i = 0; i < n; i++){
            hs.add(uf.find(i));
        }
        System.out.println(hs.size());
    }
}

class UnionFindTree {
	int[] parent;
	int[] rank;
	int[] size;
	
	public UnionFindTree(int size) {
		this.parent = new int[size];
		this.rank = new int[size];
		this.size = new int[size];
		
		for (int i = 0; i < size; i++) {
			makeSet(i);
		}
	}
	
	public void makeSet(int i) {
		parent[i] = i;
		rank[i] = 0;
		size[i] = 1;
	}
	
	public void union(int x, int y) {
		int xRoot = find(x);
		int yRoot = find(y);
		if (xRoot > yRoot) { 
			parent[yRoot] = xRoot;
			size[xRoot] += size[yRoot];
		} else if(xRoot < yRoot) {
			parent[xRoot] = yRoot;
			size[yRoot] += size[xRoot];
		} else if (xRoot != yRoot){
			parent[yRoot] = xRoot;
			size[xRoot] += size[yRoot];
			rank[xRoot]++; 
		}
	}
	
	public int find(int i) {
		if (i != parent[i]) {
			parent[i] = find(parent[i]);
		}
		return parent[i];
	}
	
	public boolean same(int x, int y) {
		return find(x) == find(y);
	}

	public int size(int x) {
		return size[find(x)];
	}
}

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;
    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken("\n");
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public double nextDouble() {
         return Double.parseDouble(next());
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    } 
}
