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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>

using namespace std;

#define ASIZE 100010

void dfs(const int& i, vector<int> (&ad)[ASIZE], vector<bool>& visited) {
    if ( visited[i] ) return;
    visited[i] = true;
    for (auto v : ad[i] ) {
        dfs(v, ad, visited);
    }
}

int main() {
    int n, m;
    cin >> n >> m;
    vector < int > ad [ASIZE];
    vector <bool> visited (ASIZE, false);
    int ans = 0;

    for (int i=0; i<m; i++) {
        int x, y, z;
        cin >> x >> y >> z;
        ad[x].push_back(y);
        ad[y].push_back(x);
    }
    

    for (int i=1; i<=n; i++) {
        if ( visited[i] ) continue;
        ans++;
        dfs(i, ad, visited);
    }

    cout << ans << endl;
}