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 {
    static Scanner scanner = new Scanner(System.in);

    public static void main(String[]$) {
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        UnionFind unionFind = new UnionFind(n);
        for (int i = 0; i < m; i++) {
            unionFind.unite(scanner.nextInt() - 1, scanner.nextInt() - 1);
            scanner.next();
        }

        System.out.println(unionFind.parents);
    }

    static class UnionFind {
        int parents;
        private int[] parent;

        UnionFind(int n) {
            parents = n;
            parent = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
        }

        int root(int n) {
            while (parent[n] != n) {
                n = parent[n] = parent[parent[n]];
            }
            return n;
        }

        void unite(int x, int y) {
            x = root(x);
            y = root(y);
            if (x != y) {
                parents--;
                parent[x] = y;
            }
        }

        boolean same(int x, int y) {
            return root(x) == root(y);
        }
    }
}