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>
using namespace std;
using Graph = vector<vector<int>>;

// 深さ優先探索
vector<bool> seen;
void dfs(const Graph &G, int v) {
    seen[v] = true;
    for (auto next_v : G[v]) { 
        if (seen[next_v]) continue;
        dfs(G, next_v); // 再帰的に探索
    }
}

int main() {
    // 頂点数と辺数
    int N, M; cin >> N >> M;

    // グラフ入力受取
    Graph G(N);
    for (int i = 0; i < M; ++i) {
        int a, b,c;
        cin >> a >> b >> c;
        G[a-1].push_back(b-1);
        G[b-1].push_back(a-1);
    }

    // 全頂点が訪問済みになるまで探索
    int count = 0;
    seen.assign(N, false);
    for (int v = 0; v < N; ++v) {
        if (seen[v]) continue; // v が探索済みだったらスルー
        dfs(G, v); // v が未探索なら v を始点とした DFS を行う
        ++count;
    }
    cout << count << endl;
}