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>

#define maxn 101111
int fa[maxn];
int vis[maxn];
int find_father(int x)
{
    if(x == fa[x])return x;
    return fa[x] = find_father(fa[x]);
}
void uni(int x,int y)
{
    int fx = find_father(x);
    int fy = find_father(y);
    if(fy!=fx)
    {
        fa[fx] = fy;
    }
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i)fa[i] = i;
    for(int i=1;i<=m;++i)
    {
        int x,y,w;
        scanf("%d%d%d",&x,&y,&w);
        uni(x,y);
    }
    int ans = 0;
    for(int i=1;i<=n;++i)
    {
        if(!vis[find_father(i)])
        {
            vis[find_father(i)]=1;
            ans++;
        }
    }
    printf("%d\n",ans);
}