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)

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// ░░░░░░███████ ]▄▄▄▄▄▄▄▄▃░░░▃░░░░ ▃░
// ▂▄▅█████████▅▄▃▂░░░░░░░░░░░░░░░░░
// [███████████████████].░░░░░░░░░░░░░░
// ◥⊙▲⊙▲⊙▲⊙▲⊙▲⊙▲⊙◤...░░░░░░░░░░░░

//PointBlank's code (⌐■_■)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
 
#define repp(I, A, B) for(int I = A; I <= B; I++)
#define rep(I, A, B) for(int I = A; I < B; I++)
#define rrep(I, B, A) for(int I = B; I >= A; I--)
#define pb emplace_back
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define LB lower_bound
#define UB upper_bound
#define MP make_pair
#define mem(A, B) memset(A, B, sizeof(A));
#define mem0(A) memset(A, 0, sizeof(A));
#define quickio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define debug(this) cerr<<"> "<<#this<<" : "<<this<<"\n"
#define bitsf(n) __builtin_popcount(n)
#define in_edges(M) repp(I, 1, M){ll A, B; cin >> A >> B; v[A].pb(B), v[B].pb(A);}
#define in_edges2(M) repp(I, 1, M){ll A, B; cin >> A >> B; v[A].pb(B);}
#define endl "\n"
ll power(ll x, ll y); 

const ll MOD = 1000000007;
const int MX = 1e5 + 6;

vector<int> v[MX];
bool vis[MX];
int X[MX], Y[MX], trash[MX], res;

void dfs(int s){
	vis[s] = 1;
	for(auto i : v[s]) if(!vis[i]) dfs(i);
}
 
int main() //PointBlank's code ¯\_(ツ)_/¯
{
	int n, m;
	cin >> n >> m;
	repp(i, 1, m) cin >> X[i] >> Y[i] >> trash[i];
	repp(i, 1, m) v[X[i]].pb(Y[i]), v[Y[i]].pb(X[i]);
	repp(i, 1, n) if(!vis[i]){
		dfs(i);
		res++;
	}
	cout << res;
}

ll power(ll x, ll y)
{ 
    ll res = 1; x %= MOD;
    while(y > 0){if(y & 1) res = (res*x) % MOD; y = y>>1, x = (x*x) % MOD;} 
    return res; 
}