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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using  namespace std;

class UnionFind {
  public:

  //par[i]:parent of i. rank[i]:rank(depth) of the group with i.
  vector<int> par, rank; 

  UnionFind(int n) {
    par = vector<int>(n, -1);
    rank = vector<int>(n, 0);
  }

  int root(int x) {               //root(x):root of x
    return (par[x] < 0) ? x : par[x] = root(par[x]);
  }

  int size(int x) {               //the number of edge of the group with x
    return -par[root(x)];
  }

  bool same(int x, int y) {       //judge whether x & y is in the same group.
    return root(x) == root(y);
  }

  void unite(int x, int y) {      //unite group with x & group with y.
    x = root(x);
    y = root(y);
    if (x == y) return;

    if (rank[x] < rank[y]) swap(x, y);
    par[x] += par[y];
    par[y] = x;
    if (rank[x] == rank[y]) rank[x]++;
  }
};

int main() {
  int n, m;
  cin >> n >> m;
  UnionFind uf(n);
  rep(i, m) {
    int x, y, z;
    cin >> x >> y >> z;
    --x; --y;
    uf.unite(x, y);
  }
  int ans = 0;
  rep(i, n) if (uf.root(i) == i) ++ans;
  cout << ans << endl;
  return 0;
}