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<algorithm>
#include<vector>
#include<queue>
#include<string>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> P;
vector<P> node[100005];
bool used[100005];
int ans[100005]={0};
void dfs(int pos){
  for(int i=0;i<node[pos].size();i++){
    int to=node[pos][i].first;
    if(!used[to]){
      used[to]=true;
      dfs(to);
    }
  }
}
int main(){
  ll n,m;
  cin>>n>>m;
  for(int i=0;i<m;i++){
    ll x,y,z;
    cin>>x>>y>>z;
    z=z%2;
    node[x].push_back(P(y,z));
    node[y].push_back(P(x,z));
  }
  for(int i=1;i<=n;i++) used[i]=false;
  ll ans=0;
  for(int i=1;i<=n;i++){
    if(!used[i]) {
      used[i]=true;
      dfs(i);
      ans++;
    }
  }
  cout<<ans<<endl;
}