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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int main() {
	int n, m;
	int a, b, c;
	int N[100000];
	scanf("%d %d", &n, &m);
	int i;
	int tp;
	int count = 0;
	int groups = 0;
	for (i = 0; i < n; i++) N[i] = 0;
	for (i = 0; i < m; i++) {
		scanf("%d %d %d", &a, &b, &c);
        a=a-1;
        b=b-1;
		if (N[a] == 0 && N[b] == 0) {
			count++;
			groups++;
			N[a] = groups;
			N[b] = groups;
		}
		else if (N[a] == 0 && N[b] != 0) {
			count++;
			N[a] = N[b];
		}
		else if (N[a] != 0 && N[b] == 0) {
			count++;
			N[b] = N[a];
		}
		else if( N[a] != N[b] ){
			count++;
			tp = N[b];
			for (i = 0; i < n; i++) {
				if (N[i] == tp) N[i] = N[a];
			}
		}
	}
	printf("%d\n", n-count);

	return 0;
}