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: Wrong Answer)

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

//ABC126E
int main(void) {
  int n, m, k, i, c=1, tmp;
  scanf("%d %d", &n, &m);
  int x[m], y[m], z[n];
  for(i=0; i < m; i++){
    scanf(" %d %d %d", &x[i], &y[i], &tmp);
  }
  for(i=0; i < n; i++)
    z[i]=0;
  for(i=0; i < m; i++){
    if(z[x[i] - 1] ==0 && z[y[i] - 1]==0) z[x[i] - 1]=c, z[y[i] - 1]=c, c++;
    else if(z[x[i - 1]] != 0) z[y[i] - 1]=z[x[i] - 1];
    else z[x[i] - 1]=z[y[i] - 1];
   // printf("z=%d %d\n", z[x[i] - 1], z[y[i] - 1]);
  }

  for(i=0; i < n; i++){
    if(z[i] == 0) c++;
    //printf("%d\n", z[i]);
  }
  printf("%d", c - 1);
}