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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

n,m=map(int, input().split())

tree=[[] for _ in range(n)]
for i in range(m):
    x,y,w = map(int,input().split())
    tree[x-1].append(y-1)
    tree[y-1].append(x-1)

singles = [ i for i in range(n) if tree[i]==[]]

visited=[False]*n
for i in singles:
    visited[i] = True
ans=len(singles)

while False in visited:
    ans+=1
    restart = visited.index(False)
    visited[restart]=True
    if tree[restart]==[]:
        continue
    q = [(restart,tree[restart][0])]
    while q:
        parent,x = q.pop()
        visited[x]=True
        for y in tree[x]:
            if visited[y]:
                continue
            q.append((x,y))
print(ans)