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>

using namespace std;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int N,M; cin>>N>>M;

  vector<int> parent(N);
  iota(parent.begin(), parent.end(), 0);
  auto root = [f=[&](auto g, int n) {
    if (parent[n] == n)
      return n;
    return parent[n] = g(g, parent[n]);
  }] (int n) {
    return f(f, n);
  };

  for (int i = 0; i < M; i++) {
    int x,y,z;cin>>x>>y>>z;
    x--;y--;
    auto rl = root(x);
    auto rr = root(y);
    if (rl != rr) {
      parent[rl] = rr;
    }
  }

  std::set<int> graphs;
  for (int i = 0; i < N; i++) {
    graphs.insert(root(i));
  }
  cout << graphs.size() << endl;

  return 0;
}