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 Python (3.8.2) (judge verdict: Time Limit Exceeded)

from collections import deque
N,M=list(map(int,input().split()))
even_list=[list(map(int,input().split())) for _ in range(M)]
#リストVのN番目には頂点Nがあり、そこには頂点Nとつながっている頂点が入っている。
    
#seen[i]は、頂点iを訪問済みの場合はTrue、未訪問の場合はFalse
seen=[False for i in range(N+1)]


count=0
graph=[[] for _ in range(N+1)]
for i in even_list:
    graph[i[0]].append(i[1])
    graph[i[1]].append(i[0])

def StackBFS(s):
    #graphのN番目には頂点Nがあり、そこには頂点Nとつながっている頂点が入っている。
    S=deque([s])
    C=set()
    
    while S:
        v=S.pop()
        seen[v]=True #訪問済みの頂点をTrueにする
        if v not in C:
            C=C|{v}
            for i in graph[v]:
                S.append(i)

for i in range(1,N+1):
    if seen[i]==False:
        StackBFS(i)
        count+=1

print(count)