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 <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

int main(){
  int N,M;
  cin >> N >> M;
  int a[N];
  rep(i,N) a[i]=-1;
  int cnt=0;
  vector<vector<int>> vv(N);
  rep(i,M){
    int x,y,z;
    cin >> x >> y >> z;
    x--;y--;
    vv[x].push_back(y);
    vv[y].push_back(x);
  }
  queue<int> q;
  int i=0;
  int f=0;
  while(i<N){
    if(a[i]==-1){
      q.push(i);
      cnt++;
      a[i]=cnt;
      while(!q.empty()){
        int n=q.front();
        q.pop();
        for(int x:vv[n]){
          if(a[x]==-1){
            q.push(x);
            a[x]=cnt;
          }
        }
      }
    }
    i++;
  }
  cout << cnt << endl;
  return 0;
}