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>
#include<iomanip>
#include<numeric>

using namespace std;
using ll = long long;
constexpr int mo = 1e9+7;
constexpr int  mod = mo;
constexpr ll inf = 1ll<<62;
class UF{
    vector<int> par;
    public:
    UF(int n):par(n){iota(par.begin(),par.end(),0);}
    int operator[](int x){ return (par[x] == x)?x:par[x] = UF::operator[](par[x]);}
    void operator()(int x,int y){
        x = operator[](x);y = operator[](y);
        if(x == y)return;
        if(x > y)swap(x,y);
        par[y] = x;
    }
};
int main(){
    int n,m;
    cin>>n>>m;
    UF uf(n+1);
    for(int i=0;i<m;++i){
        int x,y,z;
        cin>>x>>y>>z;
        x--;
        y--;
        uf(x,y);
    }
    map<int ,int> mp;
    int count = 0;
    for(int i=0;i<n;++i){
        if(!mp.count(uf[i])){
            count++;
            mp[uf[i]]++;
        }
    }
    cout << count << endl;
    return 0;
}