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: Accepted)

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

#define LIST_ISIZE 1

typedef struct {
	int n, c, *v;
} list;

void init(list *l) {
	l->n = 0;
	l->c = LIST_ISIZE;
	l->v = malloc(l->c * sizeof *l->v);
}

void add(list *l, int v) {
	if (l->n+1 > l->c) {
		l->c *= 2;
		l->v = realloc(l->v, l->c * sizeof *l->v);
	}
	l->v[l->n++] = v;
}

typedef struct {
	list same, diff;
} node;

node graph[100000];
int paint[100000];

void dfs(int v) {
	node gc = graph[v];
	for (int i = 0; i < gc.same.n; i++) {
		if (paint[gc.same.v[i]] != -1) continue;
		paint[gc.same.v[i]] = paint[v];
		dfs(gc.same.v[i]);
	}
	for (int i = 0; i < gc.diff.n; i++) {
		if (paint[gc.diff.v[i]] != -1) continue;
		paint[gc.diff.v[i]] = !paint[v];
		dfs(gc.diff.v[i]);
	}
}

int main(void) {
	int n, m;
	scanf("%d%d", &n, &m);

	for (int i = 0; i < n; i++) {
		init(&graph[i].same);
		init(&graph[i].diff);
	}

	for (int i = 0; i < m; i++) {
		int a, b, s;
		scanf("%d%d%d", &a, &b, &s);
		a--; b--; s %= 2;
		if (s) {
			add(&graph[a].diff, b);
			add(&graph[b].diff, a);
		} else {
			add(&graph[a].same, b);
			add(&graph[b].same, a);
		}
	}

	for (int i = 0; i < n; i++) paint[i] = -1;

	int cnt = 0;
	for (int i = 0; i < n; i++) {
		if (paint[i] != -1) continue;
		paint[i] = 0;
		cnt++;
		dfs(i);
	}

	for (int i = 0; i < n; i++) {
		free(graph[i].same.v);
		free(graph[i].diff.v);
	}

	printf("%d\n", cnt);
}