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)

// :)
// "Khodaya, be man "Tagwaye setiz" biamooz ta
//  dar anbuh masuliat nalaghzam ..." -Shariati
#include <bits/stdc++.h>

using namespace std;
typedef long long int64;
typedef pair<int,int> pii;
typedef pair<int64,int> pii32;
typedef pair<int64,int64> pii64;

#define PB push_back
#define MP make_pair
#define F first
#define S second
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()

const int maxn = 1e5+10;
const int64 MO = 1e9+7;
const int64 IN = 1e9;

vector <pii> v1[2];
int par[2][maxn];
int n, m;

int root (int id, int v) {
	return ((par[id][v] < 0) ? v : par[id][v] = root(id, par[id][v]));
}

bool merge (int id, int v, int u) {
	v = root(id, v);
	u = root(id, u);
	if (v == u)
		return 0;
	par[id][v] = u;
	return 1;
}

int main () {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int x, y, z;
		cin >> x >> y >> z;
		--x; --y;
		v1[z & 1].PB(MP(x, y));
	}
	memset(par, -1, sizeof par);
	for (auto x : v1[0])
		merge(0, x.F, x.S);
	int cnt = 0;
	for (int i = 0; i < n; i++)
		if (i == root(0, i))
			cnt++;
	for (auto x : v1[1])
		if (merge(1, root(0, x.F), root(0, x.S)))
			cnt--;
	cout << cnt << "\n";
}