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++ (GCC 9.2.1) (judge verdict: Runtime Error)

// 7/4
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

struct UnionFind {
  vector<int> d;
  UnionFind(int n=0): d(n,-1) {}
  int find(int x) {
    if (d[x] < 0) return x;
    return d[x] = find(d[x]);
  }
  bool unite(int x, int y) {
    x = find(x); y = find(y);
    if (x == y) return false;
    if (d[x] > d[y]) swap(x,y);
    d[x] += d[y];
    d[y] = x;
    return true;
  }
  bool same(int x, int y) { return find(x) == find(y);}
  int size(int x) { return -d[find(x)];}
};

int main() {
    int N, M;
    cin >> N >> M;
    vector<int> X(M), Y(M), Z(M);
    rep(i, M) cin >> X[i] >> Y[i] >> Z[i];

    // UnionFind uf(2*N);
    // rep(i, M) {
    //     int x = X[i];
    //     int y = Y[i];
    //     if (Z[i]%2) {
    //         // X, Yは偶奇が等しい
    //         uf.unite(x, y);
    //         uf.unite(x+N, y+N);
    //     } else {
    //         // X, Yは偶奇が異なる
    //         uf.unite(x, y+N);
    //         uf.unite(x+N, y);
    //     }
    // }

    // set<int> st;
    // rep(i, 2 * N) {
    //     st.insert(uf.find(i));
    // }
    // cout << st.size() / 2 << endl;

    UnionFind uf(N);
    rep(i, M) {
        uf.unite(X[i], Y[i]);
    }
    set<int> st;
    rep(i, N) st.insert(uf.find(i));
    cout << st.size() << endl;
}