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: Wrong Answer)

/* ****TCIITB**** */

#include<bits/stdc++.h>
#define io          ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0); 
#define ll          long long
#define pb          push_back
#define	endl		'\n'
#define pii         pair<ll int,ll int>
#define vi          vector<ll int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll int)x.size()
#define hell        1000000007
#define rep(i,a,b)	for(ll int i=a;i<b;i++)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp          make_pair
#define time        cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
using namespace std;

#define N  100005
ll n,m;
std::vector<ll> adj[N];
std::map<ll,ll> cnt;
int main()
{
	
	io
	cin>>n>>m;
	std::map<pii,ll> vis;
	rep(i,0,m)
	{
		ll x,y,z;
		cin>>x>>y>>z;
		if(vis[{x,y}] || vis[{y,x}])
			continue;
		vis[{x,y}]++;
		vis[{y,x}]++;
		adj[x].pb(y);
		adj[y].pb(x);
	}
	std::vector<pii> tmp;
	rep(i,1,n+1)
	{
		tmp.pb({adj[i].size(),i});
	}
	sort(tmp.rbegin(), tmp.rend());
	ll ans=0;
	rep(i,0,tmp.size())
	{
		if(cnt.size()==(n))
			break;		
		ll z=tmp[i].S;

		if(cnt[z]!=0)
			continue;
		cnt[z]++;
		for(auto j:adj[z])
		{
			cnt[j]++;
			for(auto k:adj[j])
			{
				cnt[k]++;
			}
		}
		ans++;
	}
	cout<<ans;

	
}