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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

int main(void){
  int i,j,N,M,x[100005],y[100005],z[100005],a[100005]={},cost,count,h,buff;
  char str[1000000],*p;
  
  fgets(str,sizeof(str),stdin);
  p=strtok(str," \n");
  N=atoi(p);
  p=strtok(NULL," \n");
  M=atoi(p);
  
  for(i=0;i<M;i++){
    fgets(str,sizeof(str),stdin);
    p=strtok(str," \n");
    x[i]=atoi(p);
    p=strtok(NULL," \n");
    y[i]=atoi(p);
    p=strtok(NULL," \n");
    z[i]=atoi(p);
  }
  
  for(h=1;h<M/9;h=3*h+1);
  
  for(;h>0;h/=3){
    for(i=0;i<M;i++){
      j=i;
      while(h<=j && x[j-h]>x[j]){
        buff=x[j-h],x[j-h]=x[j],x[j]=buff;
        buff=y[j-h],y[j-h]=y[j],y[j]=buff;
      }
    }
  }
  
  for(i=0,cost=0;i<M;i++){
    if(a[x[i]]==0 && a[y[i]]==0){
      cost++;
      a[x[i]]=1,a[y[i]]=1;
    }else if(a[x[i]]==1){
      a[y[i]]=1;
    }else if(a[y[i]]==1){
      a[x[i]]=1;
    }
  }
  
  for(i=1,count=0;i<N+1;i++){
    if(a[i]==0){
      count++;
    }
  }
  
  cost+=count;
  
  printf("%d\n",cost);
  
  return 0;
}