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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

typedef struct List {
	struct List *next;
	int v;
} list;

int main()
{
	int i, u, w, z, N, M;
	scanf("%d %d", &N, &M);
	list **adj = (list**)malloc(sizeof(list*) * (N + 1)), *d = (list*)malloc(sizeof(list) * M * 2);
	for (i = 1; i <= N; i++) adj[i] = NULL;
	for (i = 0; i < M; i++) {
		scanf("%d %d %d", &u, &w, &z);
		d[i*2].v = w;
		d[i*2+1].v = u;
		d[i*2].next = adj[u];
		d[i*2+1].next = adj[w];
		adj[u] = &(d[i*2]);
		adj[w] = &(d[i*2+1]);
	}
	
	int k, flag[100001] = {}, q[100001], head, tail;
	list *p;
	for (i = 1, k = 0; i <= N; i++) {
		if (flag[i] != 0) continue;
		flag[i] = ++k;
		q[0] = i;
		for (head = 0, tail = 1; head < tail; head++) {
			for (p = adj[q[head]]; p != NULL; p = p->next) {
				if (flag[p->v] == 0) {
					flag[p->v] = k;
					q[tail++] = p->v;
				}
			}
		}
	}
	
	printf("%d\n", k);
	fflush(stdout);
	return 0;
}