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 <iostream>
#include <vector>
#include <queue>
#include <utility>
using namespace std;

void search(int c, vector<int> *lst, bool *v){
  
  for (int i = 0; i < (lst + c)->size(); i++) {
    
    if (v[(*(lst + c))[i]] == false) {
      v[(*(lst + c))[i]] = true;
      search((*(lst + c))[i], lst, v);
    }
  }
  return;
}

int main(void) {
  int n, m, count;
  vector<int> *lst;
  bool *visited;
  
  cin >> n >> m;
  visited = new bool[n+1];
  lst = new vector<int>[n+1];
  
  for (int i = 0; i < m; i++) {
    int u, v, z;
      
    cin >> u >> v >> z;
    lst[u].push_back(v);
    lst[v].push_back(u);
  }
  
  for (int i = 1; i < n + 1; i++) {
    if (visited[i] == false) {
      count++;
      visited[i] = true;
      search(i, lst, visited);
    }
  }
  
    
  cout << count << endl;
  
  return 0;
}