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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void)
{
    int i, ii, n, m, x[100001], y[100001], z[1], f[100001] = {}, cnt = 0;
    scanf("%d%d", &n, &m);
    for (i = 0; i < m; i++)
        scanf("%d%d%d", x + i, y + i, z);
    for (i = 0; i < m; i++)
    {
        if (f[x[i]] && !f[y[i]])
            f[y[i]] = f[x[i]];
        else if (f[y[i]] && !f[x[i]])
            f[x[i]] = f[y[i]];
        else if (!f[x[i]])
        {
            f[y[i]] = i + 1;
            f[x[i]] = i + 1;
        }
        else if (f[y[i]] != f[x[i]])
        {
            for (ii = 0; ii < n; ii++)
                if (f[ii] == f[y[i]])
                    f[ii] = f[x[i]];
        }
        else
            cnt--;

        cnt++;
    }
    printf("%d", n - cnt);
}