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 (3.4.3) (judge verdict: Wrong Answer)

N,M=map(int, input().split())

outgoing=dict()
incoming=dict()
for i in range(M):
  x,y,z=map(int, input().split())
  if x in outgoing: outgoing[x].add(y)
  else: outgoing[x]={y}
  if y in incoming: incoming[y].add(x)
  else: incoming[y]={x}
#print(outgoing)
#print(incoming)
V=set()
def dfs(v):
  if v not in V:
    V.add(v)
    """
    if v in outgoing:
      for u in outgoing[v]: dfs(u)
    """
    if v in incoming:
      for u in incoming[v]: dfs(u)
cnt=0
for i in range(1, N+1):
  if i not in V:
    cnt += 1
    dfs(i)
print(cnt)