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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdlib>
#include <iostream>
#include <stdio.h>
#include <vector>
#include <queue>


typedef struct _xyz {
    int x;
    int y;
    int z;
} xyz;


typedef struct _node {
    std::vector<int> nbr; // zero-based numbering (node_arr[nbr]
} node;


int main(void)
{
    int n, m;
    scanf("%d%d", &n, &m);
    xyz *xyz_arr = (xyz *)std::malloc(sizeof(xyz) * m);
    for (int i = 0; i < m; ++i) {
        scanf("%d%d%d", &xyz_arr[i].x, &xyz_arr[i].y, &xyz_arr[i].z);
    }

    node *node_arr = (node *)std::malloc(sizeof(node) * n);
    for (int i = 0; i < m; i++) {
        node_arr[xyz_arr[i].x - 1].nbr.push_back(xyz_arr[i].y - 1);
        node_arr[xyz_arr[i].y - 1].nbr.push_back(xyz_arr[i].x - 1);
    }

    int count;
    std::vector<bool> memo(n);
    count = 0;
    for (int i = 0; i < n; ++i) {
        if (!memo[i]) {
            /*
             * BFS
             */
            memo[i] = true;
            std::queue<int> q; // zero-based numbering
            q.push(i);
            while (!q.empty()) {
                int j = q.front();
                q.pop();
                for (auto itr = node_arr[j].nbr.begin(); itr != node_arr[j].nbr.end(); ++itr)  {
                    if (!memo[*itr]) {
                        memo[*itr] = true;
                        q.push(*itr);
                    } else {
                        continue;
                    }
                }
            }
            count++;
        } else if (node_arr[i].nbr.empty()) {
            count++;
        } else {
            continue;
        }
    }

    std::cout << count << std::endl;

    return EXIT_SUCCESS;
}