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;
typedef long long ll;
typedef string str;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int,int> vii;
typedef vector<ll,ll> vll;
typedef vector <pii> vpii;
typedef vector <pll> vpll;
typedef map <str,int> mapsi;
typedef map <str,int> :: iterator mapsitr;
typedef set <int> :: iterator sitr;
typedef set <ll> :: iterator sltr;
#define mset multiset
typedef mset <int> msi;
typedef mset <ll> msll;
typedef mset <int> :: iterator msitr;
typedef mset <ll> :: iterator msltr;
#define mp make_pair
#define pb push_back
#define pob pop_back
#define fi first
#define se second
#define fs first.second
#define ss second.second
#define ff first.first
#define sf second.first
int parent[200000],f[200000];
int findrep(int x){
    if(parent[x]==x) return parent[x];
    else{
        return findrep (parent[x]);
    }
}
void join (int a, int b){
    int repa,repb;
    repa = findrep(a);
    repb = findrep(b);
    if(repa>repb)swap(repa,repb);
    if(repa!=repb)parent[repb]=repa;
}
int n,m,ans;
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++)parent[i]=i;
    for(int i=1;i<=m;i++){
        int a,b,c;
        cin>>a>>b>>c;
        join (a,b);
    }

    for(int i=1;i<=n;i++){
        int par = findrep(i);
        if(f[par]==0){f[par]++;ans++;}
    }
    cout<<ans<<endl;
}