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.math.*;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;

public class Main {

    static final int modnum = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();

        UnionFindTree unionfind = new UnionFindTree(N);

        for (int i = 0; i < M; i++) {
            int p1 = sc.nextInt();
            int p2 = sc.nextInt();
            int length = sc.nextInt() % 2;
            unionfind.union(p1 - 1, p2 - 1);
        }

        HashMap<Integer,Integer> roots = new HashMap<Integer,Integer>();
        for (int i = 0; i < N; i++) {
            roots.put(unionfind.find(i), 1);
        }

        System.out.println(roots.size());
    }
}

class UnionFindTree {

    int[] parent; // インデックスにとノードを対応させ、そのルートノードのインデックスを格納
    int[] rank; // parentと同様に、木の高さを格納

    public UnionFindTree(int size) {
        this.parent = new int[size];
        this.rank = new int[size];

        for (int i = 0; i < size; i++) {
            makeSet(i);
        }
    }

    public void makeSet(int i) {
        parent[i] = i;
        rank[i] = 0; // 集合の高さ
    }

    public void union(int x, int y) {
        int xRoot = find(x);
        int yRoot = find(y);

        // xが属する木の方が大きい場合
        if (rank[xRoot] > rank[yRoot]) {
            parent[yRoot] = xRoot; // yの親をxに更新

        } else if (rank[xRoot] < rank[yRoot]) {
            parent[xRoot] = yRoot;

        } else if (xRoot != yRoot) {
            parent[yRoot] = xRoot;
            rank[xRoot]++; // 同じ高さの木がルートの子として着くから大きさ++;
        }
    }

    // 引数の属する木のルートのidを返す
    public int find(int i) {
        if (i != parent[i]) {
            parent[i] = find(parent[i]);
        }
        return parent[i];
    }

    // 同じ木に属しているか
    public boolean same(int x, int y) {
        return find(x) == find(y);
    }
}