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();
    List<Integer>[] edges = new ArrayList[N];
    for (int i = 0; i < N; i++) {
      edges[i] = new ArrayList<Integer>();
    }
    for (int i = 0; i < M; i++) {
      int x = sc.nextInt();
      int y = sc.nextInt();
      int w = sc.nextInt();
      edges[x-1].add(y-1);
      edges[y-1].add(x-1);
    }
    boolean[] visited = new boolean[N];
    int ans = 0;
    for (int i = 0; i < N; i++) {
      if (visited[i]) continue;
      Queue<Integer> queue = new LinkedList<Integer>();
      queue.offer(i);
      visited[i] = true;
      ans++;
      while (queue.size() > 0) {
        int v = queue.poll();
        for (int w : edges[v]) {
          if (visited[w]) continue;
          queue.offer(w);
          visited[w] = true;
        }
      }
    }
    System.out.println(ans);
  }
}