Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

3 1
1 2 1

Sample Output 1

2

You can determine all of A_1, A_2, A_3 by using the magic for the first and third cards.

Sample Input 2

6 5
1 2 1
2 3 2
1 3 3
4 5 4
5 6 5

Sample Output 2

2

Sample Input 3

100000 1
1 100000 100

Sample Output 3

99999

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {

    class UF {

        int n;
        int[] par; // 親の要素の番号
        int[] rank; // 木の高さ

        public UF(int n) {
            this.n = n;
            par = new int[n];
            rank = new int[n];
            for (int i = 0; i < n; i++) {
                par[i] = i;
                rank[i] = 0;
            }

        }

        // 木の根を求める
        int find(int x) {
            if (par[x] == x) {
                return x;
            } else {
                return par[x] = find(par[x]);
            }
        }

        // xとyの属する集合を併合
        void unite(int x, int y) {
            x = find(x);
            y = find(y);
            if (x == y) return;

            if (rank[x] < rank[y]) {
                par[x] = y;
            } else {
                par[y] = x;
                if (rank[x] == rank[y]) rank[x]++;
            }
        }

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

    void run() {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = sc.nextInt();
        int m = sc.nextInt();

        UF uf = new UF(n);

        for (int i = 0; i < m; i++) {
            int x = sc.nextInt() - 1;
            int y = sc.nextInt() - 1;
            int z = sc.nextInt();

//            if (z % 2 == 0) {
//                // xとyは同じグループ
//                uf.unite(x, y);
//            } else {
//                // ?
//            }
            uf.unite(x, y);

        }
//        debug(uf.par);
//        debug(uf.rank);

        Set<Integer> set  = new HashSet<>();
        for (int i : uf.par) {
            set.add(i);
        }
        System.out.println(set.size());

        out.flush();
    }

    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}