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)

#include<iostream>
#include<iomanip>
#include<vector>
#include<string>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<climits>
#include<fstream>

#define MOD (1000000007)
#define N_MAX 100000+10

using namespace std;

typedef long long int Int;

class UnionFind {
    private:
        int par[N_MAX];
        int rank[N_MAX];
    public:
        UnionFind(int n) {
            for(int i = 0;i < n;i++) {
                par[i] = i;
                rank[i] = 0;
            }
        }

        int root(int x) {
            return x == par[x] ? par[x] : par[x] = root(par[x]);
        }

        bool same(int x,int y) {
            return par[x] == par[y];
        }

        void merge(int x,int y) {
            x = root(x);
            y = root(y);
            if(par[x] == par[y])
                return;

            if(rank[x] < rank[y]) {
                par[x] = y;
            } else {
                par[y] = x;
                if(rank[x] == rank[y])
                    rank[x]++;
            }

            return;
        }

        int num_cls(int n) {
            map<int,bool> dic;

            for(int i = 0;i < n;i++) {
                dic[par[i]] = true;
            }

            return dic.size();
        }
};

int main(void) {
    int n,m;
    cin>>n>>m;

    UnionFind uf(n);

    for(int i = 0;i < m;i++) {
        int x,y,z;
        cin>>x>>y>>z;
        x--;
        y--;
        uf.merge(x,y);
    }

    cout<<uf.num_cls(n)<<endl;

    return 0;
}