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 vector<ll> vll;
typedef pair<ll, ll> pll;
typedef long double ld;
const ll mod = 1e9 + 7;

ll solve(void)
{
	ll n, m;
	cin >> n >> m;

	vector<vll> e(n, vll());

	for (ll i = 0; i < m; i++)
	{
		ll x, y, z;
		cin >> x >> y >> z;
		x--; y--;
		e[x].push_back(y);
		e[y].push_back(x);
	}

	ll ans = 0;
	vector<bool> visited(n, false);
	for (ll i = 0; i < n; i++)
	{
		if (visited[i]) { continue; }
		ans ++;
		visited[i] = true;

		queue<ll> q;
		q.push(i);

		while (!q.empty())
		{
			ll f = q.front();
			q.pop();
			visited[f] = true;
			for (auto i : e[f])
			{
				if (visited[i]) { continue; }
				q.push(i);
			}
		}
	}
	
	cout << ans << endl;


	return 0;
}

int main(void)
{
	while(solve()){}

	return 0;
}