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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int M = scanner.nextInt();
    Node[] nodes = new Node[N];
    for (int i = 0; i < N; i++) nodes[i] = new Node();
    for (int i = 0; i < M; i++) {
      int x = scanner.nextInt() - 1;
      int y = scanner.nextInt() - 1;
      int z = scanner.nextInt();
      nodes[x].adjacent.add(nodes[y]);
      nodes[y].adjacent.add(nodes[x]);
    }

    int count = 0;
    for (int i = 0; i < N; i++) {
      if (!nodes[i].visited) {
        count++;
        dfs(nodes[i]);
      }
    }
    System.out.println(count);
  }

  private static void dfs(Node node) {
    node.visited = true;
    for (Node n : node.adjacent) {
      if (!n.visited) dfs(n);
    }
  }

  private static class Node {
    List<Node> adjacent = new ArrayList<>();
    boolean visited;
  }
}
