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<bits/stdc++.h>
using namespace std;
int n,m;
bool visited[100005];
#define ll long long int
#define MOD 1000000007
vector<ll>grp[100005];
ll coun,ans;
void explore(ll s){
	visited[s]=true;
	//cout<<s<<endl;
	for(ll i=0;i<grp[s].size();i++){
		if(!visited[grp[s][i]]){
			explore(grp[s][i]);
		}
	}
}
void dfs(ll source){
	explore(source);
	for(ll i=1;i<=n;i++){
		if(!visited[i]){
			explore(i);
			coun++;
			
		}
		
	}
}
int main(){
	ll t=1;
	while(t--){
		for(int i=0;i<100005;i++)
			grp[i].clear();
		coun=1;
		cin>>n>>m;
		for(ll i=0;i<=n;i++){
			visited[i]=false;
		}
		for(ll i=0;i<m;i++){
			ll u,v,w;
			cin>>u>>v>>w;
			grp[u].push_back(v);
			grp[v].push_back(u);
		}
		ll source=1;
		dfs(source);
		cout<<coun<<endl;
	}
}