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 <algorithm>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <random>
#include <chrono>
#include <queue>
#include <ctime>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
#define fs first
#define sc second
#define INF 1000000000
#define MOD 1000000007
#define EPS 0.00000001


int N, M;
vector<vector<int>> E;
vector<bool> visited;
vector<bool> tojo;


void dfs(int x){
    visited[x] = true;
    for(int i=0; i<E[x].size(); i++){
        if(!visited[E[x][i]]) dfs(E[x][i]);
    }
    return;
}


int main() {
    cin >> N >> M;
    E.resize(N);
    visited.resize(N);
    tojo.resize(N);
    for(int i=0; i<M; i++){
        int x, y, z; cin >> x >> y >> z;
        E[x-1].push_back(y-1);
        E[y-1].push_back(x-1);
        tojo[x-1] = true;
        tojo[y-1] = true;
    }
    
    int cnt = 0;
    for(int i=0; i<N; i++){
        if(tojo[i] && !visited[i]){
            dfs(i);
            cnt++;
        }
    }

    for(int i=0; i<N; i++){
        if(!tojo[i]) cnt++;
    }
    cout << cnt << endl;
    return 0;
}