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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <string>

using namespace std;



int main()
{
    int n, m;
    cin >> n >> m;
    vector<int> a(n, 0);
    int c = 0;

    for (int i = 0; i < m; i++)
    {
        int x, y, z;
        cin >> x >> y >> z;
        z %= 2;
        if (i == 0)
        {
            a[x] = 1;
        }
        if (a[x] != 0 && a[y] != 0)
        {
            continue;
        }
        else
        {
            if (a[x] == 0 && a[y] == 0)
            {
                c++;
            }
            if (z == 0)
            {
                if (a[x] != 0)
                {
                    a[y] = a[x];
                }
                else
                {
                    a[x] = a[y];
                }
            }
            else
            {
                if (a[x] != 0)
                {
                    a[y] = 3 - a[x];
                }
                else
                {
                    a[x] = 3 - a[y];
                }
            }
        }
    }

    for (int i = 0; i < n; i++)
    {
        if (a[i] == 0)
        {
            c++;
        }
    }
    
    cout << c << endl;
    return 0;
}