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 <iostream>
#include <string>
#include <cstdlib>
#include <vector>
#include <sstream>
#include <set>
#include <tuple>
#include <algorithm>
#include <memory>
#include <numeric>
#include <queue>
#include <functional>
#include <cmath>
#include <cstring>
#include <map>
#include <unordered_map>
#include <sstream>
#include <iomanip>

using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;

const ll INF = 1LL << 60;
const ll MOD = 1000000007;

// https://qiita.com/ofutonfuton/items/c17dfd33fc542c222396
struct UnionFind {
	vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

	UnionFind(ll N) : par(N) { //最初は全てが根であるとして初期化
		for (ll i = 0; i < N; i++) par[i] = i;
	}

	ll root(ll x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(ll x, ll y) { // xとyの木を併合
		ll rx = root(x); //xの根をrx
		ll ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}

	bool same(ll x, ll y) { // 2つのデータx, yが属する木が同じならtrueを返す
		ll rx = root(x);
		ll ry = root(y);
		return rx == ry;
	}
};

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

	ll N, M;
	cin >> N >> M;

	vector<ll> X(M), Y(M), Z(M);

	for (ll i = 0; i < M; ++i)
	{
		cin >> X[i];
		cin >> Y[i];
		cin >> Z[i];
	}

	UnionFind tree(N);
	for (ll i = 0; i < M; ++i)
	{
		tree.unite(X[i]-1, Y[i]-1);
	}

	ll ans = 0;

	vector<bool> checked(N, false);

	set<ll> roots;

	for (ll i = 0; i < N; ++i)
	{
		roots.insert(tree.root(i));
	}

	cout << roots.size() << endl;

	//system("pause");
}