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 <string>
using namespace std;

int main(void){
	int N,M;
	cin >> N >> M;
	bool b[N+1]={0};
	int need_to_open_to_know=0;
	int known_card=0;
	for(int i = 0;i<M;i++){
		int X,Y,Z;
		cin >> X >> Y >> Z;
		if(b[X]==0 && b[Y]==0){
			need_to_open_to_know+=1;
			b[X]=1;
			b[Y]=1;
			known_card+=2;
		}
		if(b[X]==0 && b[Y]==1){
			b[X]=1;
			b[Y]=1;
			known_card+=1;
		}		
		if(b[X]==1 && b[Y]==0){
			b[X]=1;
			b[Y]=1;
			known_card+=1;
		}			
	}
	cout << (N - known_card)+need_to_open_to_know;
	return 0;
}