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>

int asc(const void *a, const void *b) {
	return *(int *)a - *(int *)b;
}

int find(int x, int par[]){
	if (par[x] == x)return x;
	else return par[x] = find(par[x], par);
}

void unite(int x, int y, int par[]){
	int rx = find(x, par);
	int ry = find(y, par);
	if (rx != ry)par[ry] = rx;
}

int main(void){
	int n, m;
	scanf("%d %d", &n, &m);
	int par[100001];
	int i;
	for (i = 1; i <= n; i++){
		par[i] = i;
	}
	for (i = 0; i < m; i++){
		int x, y, z;
		scanf("%d %d %d", &x, &y, &z);
		unite(x, y, par);
	}
	for (i = 1; i <= n; i++){
		par[i] = find(i, par);
	}
	qsort(par, n + 1, sizeof(int), asc);
	int now = 0;
	int count = 0;
	for (i = 1; i <= n; i++){
		if (par[i] != now){
			count++;
			now = par[i];
		}
	}
	printf("%d\n", count);
		return 0;
}