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: Accepted)

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
 
typedef pair<double, double> p;

#define MOD 1000000007

class UnionFind {
private:
	vector<ll> parents;
	vector<ll> ranks;

public:
	vector<ll> sizes;

	UnionFind(ll s): parents(s), ranks(s), sizes(s, 1) {
		for(ll i = 0; i < s; i++) parents[i] = i;
	}

	int find(ll x) {
		if(parents[x] == x) return x;
		else return parents[x] = find(parents[x]);
	}

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

		if(ranks[x] < ranks[y]) {
			parents[x] = y;
			sizes[y] += sizes[x];
		} else {
			parents[y] = x;
			sizes[x] += sizes[y];
			if(ranks[x] == ranks[y]) ranks[x]++;
		}
	}

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

int main() {
	cin.tie(0);
    ios::sync_with_stdio(false);

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

    UnionFind uf(N);

    for(int i = 0; i < M; i++) {
        int x, y, z;
        cin >> x >> y >> z;
        x--;
        y--;
        uf.unite(x, y);
    }

    vector<bool> c(N);

    int ans = 0;

    for(int i = 0; i < N; i++) {
        int x = uf.find(i);
        if(c[x]) continue;
        ans++;
        c[x] = true;
    }

    cout << ans << endl;
}