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 <vector>
#include <algorithm>
#include <stack>
#include <queue>
#include <list>
#include <string>
#include <numeric>
#include <math.h>
#include <set>
#include <map>
#include <fstream>
#include <sstream>
#include <iterator>
#include <complex>
using namespace std;

int main()
{
	int N,M;
	cin>>N>>M;

	vector<bool> isvalid(N,true);
	vector<pair<int,int> > query(M);
	for(int i=0; i<M; i++)
	{
		int x,y,z;
		cin>>x>>y>>z;
		x--;
		y--;
		query[i]=make_pair(x,y);	
	}
	sort(query.begin(),query.end());
	for(int i=0; i<M; i++)
	{
		int x = query[i].first;
		int y = query[i].second;
		if( (isvalid[x]||isvalid[y]) && !(isvalid[x]&&isvalid[y]))
		{
			isvalid[x] = false;
			isvalid[y] = false;
		}

		//どっちもvalid
		if(isvalid[x]&&isvalid[y])
		{
			isvalid[y] = false;
		}
	}

	int sum = 0;
	for(int i=0; i<N; i++)
	{
		if(isvalid[i])
		{
			sum++;
		}
	}
	cout<<sum<<endl;
	return 0;
}