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.Scanner;

public class Main {
    public static void main(String args[]) {
        try (Scanner in = new Scanner(System.in)) {
            int N = in.nextInt();
            int M = in.nextInt();
            int[] X = new int[M];
            int[] Y = new int[M];
            int[] Z = new int[M];
            for (int i = 0; i < M; i++) {
                X[i] = in.nextInt() - 1;
                Y[i] = in.nextInt() - 1;
                Z[i] = in.nextInt() & 1;
            }

            UnionFind uf = new UnionFind();
            uf.init(N);
            for (int i = 0; i < M; i++) {
                uf.unite(X[i], Y[i]);
            }
            int count2 = 0;
            boolean[] used = new boolean[N];
            for (int i = 0; i < N; i++) {
                if (used[uf.find(i)]) {
                    continue;
                }
                used[uf.find(i)] = true;
                count2++;
            }

            System.out.println(count2);

        }
    }
}

class UnionFind {
    private int[] par;
    private int[] rank;
    private int[] size;

    public void init(int n) {
        par = new int[n];
        rank = new int[n];
        size = new int[n];
        for (int i = 0; i < n; i++) {
            par[i] = i;
            rank[i] = 0;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (par[x] == x) {
            return x;
        } else {
            par[x] = find(par[x]);
            return par[x];
        }
    }

    public int size(int x) {
        return size[find(x)];
    }

    public void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (rank[x] < rank[y]) {
            par[x] = y;
            size[x] += size[y];
            size[y] = size[x];
        } else {
            par[y] = x;
            if (rank[x] == rank[y]) {
                rank[x]++;
            }
            size[y] += size[x];
            size[x] = size[y];
        }
    }

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