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 <cassert>
#include <climits>
#include <bitset>
#include <stack>
#include <queue>
#include <iomanip>
#include <limits>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string.h>
#include <tuple>
#include <time.h>
#include <random>

using namespace std;

typedef long long ll;
typedef pair<ll,ll> P;

long long int INF = 3e18;
double Pi = 3.1415926535897932384626;

vector<ll> G[500005];
vector<P> tree[500010];
priority_queue <ll> pql;
priority_queue <P> pqp;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop

int dx[8]={1,0,-1,0,1,1,-1,-1};
int dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "DRUL";
//ll bit[500005];
//↓,→,↑,←


#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<(x)<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define re(i,n) for(i=0;i<n;i++);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define deba(x) cout<< #x << " = " << x <<endl


ll mod = 1000000007;
ll rui(ll number1,ll number2){

    if(number2 == 0){
        return 1;
    }else{
        ll number3 = rui(number1,number2 / 2);
        number3 *= number3;
        number3 %= mod;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod;
        }
        return number3;
    }
}
ll gcd(ll number1,ll number2){

    if(number1 > number2){
        swap(number1,number2);
    }
    if(number1 == 0 || number1 == number2){
        return number2;
    }else{
        return gcd(number2 % number1,number1);
    }
}
void YES(bool condition){

    if(condition){
        p("YES");
    }else{
        p("NO");
    }
    return;}
void Yes(bool condition){

    if(condition){
        p("Yes");
    }else{
        p("No");
    }
    return;}

/*
void c3_init(){
    fact[0] = rfact[0] = 1;
    for(ll i=1; i<=800000; i++){
        fact[i] = (fact[i-1]*i) % mod;
    }
	rfact[800000] = rui(fact[800000],mod - 2);
	for(ll i=800000; i>=1; i--){
	   rfact[i-1] = rfact[i] * i;
	   rfact[i-1] %= mod;
	}
	return;}
ll c3(ll n,ll r){
    return (((fact[n] * rfact[r]) % mod) * rfact[n-r]) % mod;}
*/

ll n,m,num,sum,ans,a,b,c,d,e,g,h,w,i,j,k,q,l,r;
ll x[500005],y[500005],z[500005];
bool used[500005];
char s[500005],t[500005];
bool flag,dame;

void dfs(ll node){
	used[node] = true;
	for(int i=0;i<G[node].size();i++){
		ll newnode = G[node][i];
		if(used[newnode])continue;
		dfs(newnode);
	}
	return;
}


int main(){
	cin >> n >> m;
	for(i=0;i<m;i++){
		cin >> a >> b >> c;
		G[a].pb(b);
		G[b].pb(a);
	}
	for(i=1;i<=n;i++){
		if(!used[i]){
			ans++;
			dfs(i);
		}
	}
	p(ans);
	return 0;
}