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> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <math.h>
#include <cmath>

using namespace std;

#define int long long
#define endl "\n"
#define all(v)  v.begin(),v.end()
#define fir first
#define sec second
#define m_p make_pair
#define rep(i,n)  for(int i=0; i<(int) (n); i++)


const double pai = 3.1415926535897;
const int mod = 1000000007;
const int INF = 2000000015;
const int MAX = 510000;
const int MOD = 1000000007;
long long fac[MAX], finv[MAX], inv[MAX];


int jousu(int x00, int y00) {
	int z00 = 1;
	for (int i = 0; i < y00; i++) {
		z00 *= x00;
	}
	return z00;
}
int keta(int x00) {
	int z00 = x00;
	int w00 = 0;
	while (z00 != 0) {
		z00 /= 10;
		w00++;
	}
	return w00;
}
int krag(double xx) {
	return INF - (int)(INF - xx);
}
//x未満の要素数を返す二分探索関数
int b_s(vector<int>& vec, int xx) {
	int t = -1;
	int f = vec.size();
	while (f - t > 1) {
		int m = (t + f) / 2;
		if (vec[m] < xx) {
			t = m;
		}
		else {
			f = m;
		}
	}
	return f;
}
template<typename T>
void vecout(vector<T>& vec) { for (T t : vec) cout << t << " "; cout << endl; }
template<typename TT>
void vecin(vector<TT>& vec) { for (TT t : vec) cin >> t; }


// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}
//n個を1個以上のx組のグループに分ける重複組み合わせはcom(n-1,x-1)
//グループが0個でもいいときはnにxを足す
// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
int modinv(int a, int m) {
	int b = m, u = 1, v = 0;
	while (b) {
		int t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
int modpow(int a, int n) {
	int res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

int gcd(int xx, int yy) {
	int p = xx;
	int q = yy;
	if (q > p)swap(p, q);
	while (p % q != 0) {
		p %= q;
		swap(p, q);
	}
	return q;
}
int lcm(int xx, int yy) {
	return xx * yy / gcd(xx, yy);
}
bool prime(int xx) {
	if (xx == 1) {
		return 0;
	}
	for (int i = 2; i * i <= xx; i++) {
		if (xx % i == 0) {
			return 0;
		}
	}
	return 1;
}

vector<int>vec(100004);
int u_f(int xx) {
	if (vec[xx] == xx) {
		return xx;
	}
	vec[xx] = u_f(vec[xx]);
	return vec[xx];
}

signed main() {
	int n, m, a, b, c;
	cin >> n >> m;
	rep(i, n) {
		vec[i] = i;
	}
	rep(i, m) {
		cin >> a >> b >> c;
		a--;
		b--;
		vec[max(u_f(a), u_f(b))] = min(u_f(a), u_f(b));
	}
	set<int>se;
	rep(i, n) {
		se.insert(u_f(i));
	}
	cout << se.size() << endl;
}