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: Time Limit Exceeded)

#include<stdio.h>

int c=1;
int vi[100000]={0};
int list[100000][1000];
int ll[100000]={0};

int tanff(int v){
	int i,f=0;
	vi[v]=c;
	for(i=0;i<ll[v];i++){
		if(vi[list[v][i]]==0){
			tanff(list[v][i]);
		}
	}
	return 1;
}

int main(){
	int n,m,i;
	int xyz[100000][3];
	scanf("%d%d",&n,&m);
	for(i=0;i<m;i++){
		scanf("%d%d%d",&xyz[i][0],&xyz[i][1],&xyz[i][2]);
		xyz[i][0]--;
		xyz[i][1]--;
	}
	for(i=0;i<m;i++){
		list[xyz[i][0]][ll[xyz[i][0]]++]=xyz[i][1];
		list[xyz[i][1]][ll[xyz[i][1]]++]=xyz[i][0];
	}
	int tmp=0;
	for(c=1;;c++){
		for(i=0;i<n;i++){
			if(vi[i]==0){
				tanff(i);
			}
		}
	}
	printf("%d\n",c);
	return 0;
}