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;
typedef long long ll;
typedef vector<int> vi;

const ll INF = 1LL << 60;
const int MOD = 1e9 + 7;

#define pb push_back
#define mp make_pair
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define REPR(i, b, e) for (int i = (b); i <= (e); i++)
#define DEBUG(x) cout << #x << ": " << (x) << endl
#define DEBUGA(a) cout << #a << ": "; for (const auto& v: (a)) cout << v << endl

struct UnionFind {
    vi par, rank;

    UnionFind(int n) {
        par.resize(n); rank.resize(n, 0);
        REP(i, n) par[i] = i;
    }

    int find(int x) {
        if (par[x] == x) return x;
        return par[x] = find(par[x]);
    }

    void unite(int x, int y) {
        x = find(x); y = find(y);
        if (x == y) return;

        if (rank[x] < rank[y]) {
            par[x] = y;
        } else {
            par[y] = x;
            if (rank[x] == rank[y]) rank[x]++;
        }
    }

    bool same(int x, int y) {
        return find(x) == find(y);
    }
};

int N, M;
vi X, Y, Z;
int main() {
    cin >> N >> M;
    X.resize(M); Y.resize(M); Z.resize(M);
    REP(i, M) cin >> X[i] >> Y[i] >> Z[i];

    UnionFind uf(N);

    REP(i, M) {
        uf.unite(X[i] - 1, Y[i] - 1);
    }
    set<int> s;
    REP(i, N) s.insert(uf.find(i));

    int ans = s.size();
    cout << ans << endl;

    return 0;
}