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;

class Main {
    public static class UnionFind {

        private int[] parent;

        public UnionFind(int n) {
            // 親の番号を格納する。
            // 自分自身が親だった場合は-(その集合のサイズ)を格納する。
            this.parent = new int[n];

            for (int i = 0; i < n; i++) {
                // 最初の時点では親は自分自身
                // 集合のサイズは１
                this.parent[i] = -1;
            }
        }

        // aがどのグループに属しているか調べる。
        public int root(int a) {
            if (this.parent[a] < 0) {
                return a;
            } else {
                // 再帰的に親をたどる。
                return this.parent[a] = this.root(this.parent[a]);
            }
        }

        // aの集合のサイズを返す
        public int size(int a) {
            return -this.parent[this.root(a)];
        }

        // aとbをつなぐ。
        // 大きい方(a)に小さい方(b)をつなぐ感じ。
        public boolean connect(int a, int b) {
            // 親同士をつなぐ。
            a = this.root(a);
            b = this.root(b);

            // すでにくっついている場合
            if (a == b) {
                return false;
            }

            if (this.size(a) < this.size(b)) {
                int r = a;
                a = b;
                b = r;
            }

            this.parent[a] += this.parent[b];
            this.parent[b] = a;

            return true;
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int M = scanner.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] = scanner.nextInt() - 1;
            Y[i] = scanner.nextInt() - 1;
            Z[i] = scanner.nextInt() - 1;
        }
        UnionFind find = new UnionFind(N);
        for (int i = 0; i < M; i++) {
            find.connect(X[i], Y[i]);
        }
        boolean[] ans = new boolean[N];
        for (int i = 0; i < N; i++) {
            ans[find.root(i)] = true;
        }
        int cnt = 0;
        for (boolean b : ans) {
            if (b) {
                cnt++;
            }
        }
        System.out.println(cnt);

    }
}
