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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		
		DisjointSet dj = new DisjointSet(n * 2);
		
		for(int i = 0; i < m; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			int z = sc.nextInt();
			
			if(z % 2 == 0) {
				dj.union(x, y);
				dj.union(x + n, y + n);
			}
			else {
				dj.union(x, y + n);
				dj.union(x + n, y);
			}
		}
//		dj.printPArray();
		System.out.println(dj.numOfGroup()/2);
	}

}

class DisjointSet {
	private int n;
	private int[] p;
	private int[] rank;
	//各グループの要素数を管理する
	private int[] num;
	//グループ数を管理する
	private int numOfGroup;

	public DisjointSet(int n){
		this.n = n;

		p = new int[n + 1];
		rank = new int[n + 1];

		num = new int[n + 1];
		
		Arrays.fill(num, 1);
		
		for(int i = 1; i <= n; i++){
			makeSet(i);
		}
		
		numOfGroup = n;
	}

	private void makeSet(int x){
		p[x] = x;
		rank[x] = 0;
	}

	public void union(int x, int y){
		int p1 = findSet(x), p2 = findSet(y);
		int num1 = num[p1];
		int num2 = num[p2];
		link (findSet(x), findSet(y));
		
		if(p1 != p2) {
			num[p1] = num[p2] = num1 + num2;
			numOfGroup--;
		}
	}
	
	public int amountOfOwnGroup(int x) {
		return num[findSet(x)];
	}

	public int numOfGroup() {
		return numOfGroup;
	}
	private int findSet(int x){
		if(x != p[x]){
			p[x] = findSet( p[x]);
		}
		return p[x];
	}

	public boolean isSameSet(int x, int y){
		return findSet(x) == findSet(y);
	}

	private void link(int x, int y){
		if(rank[x] > rank[y]){
			p[y] = x;
		}
		else {
			p[x] = y;
			if(rank[x] == rank[y]){
				rank[y]++;
			}
		}
	}
	
	//for debug
	public void printPArray() {
		for(int i = 1; i <= n; i++) {
			System.out.print(p[i]+" ");
		}
		System.out.println();
	}
}
