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.*;

public class Main {
	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		UnionFindTree uft = new UnionFindTree(n);
		for (int i = 0; i < m; i++) {
		    uft.unite(sc.nextInt() - 1, sc.nextInt() - 1);
		    sc.nextInt();
		}
		System.out.print(uft.getCount());
	}
	
	static class UnionFindTree {
	    int[] parents;
	    
	    public UnionFindTree(int size) {
	        parents = new int[size];
	        for (int i = 0; i < size; i++) {
	            parents[i] = i;
	        }
	    }
	    
	    public int find(int x) {
	        if (parents[x] == x) {
	            return x;
	        } else {
	            return parents[x] = find(parents[x]);
	        }
	    }
	    
	    public void unite (int x, int y) {
	        int xx = find(x);
	        int yy = find(y);
	        if (xx == yy) {
	            return;
	        }
	        parents[xx] = yy;
        }

	    public int getCount() {
	        int count = 0;
	        for (int i = 0; i < parents.length; i++) {
	            if (parents[i] == i) {
	                count++;
	            }
	        }
	        return count;
	        
	    }
	}
}
