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)

from collections import defaultdict
import sys,heapq,bisect,math,itertools,string,queue,datetime
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().split())

def Find(x):    #xの根を返す
    global table

    if table[x] == x:
        return x
    else:
        table[x] = Find(table[x])    #親の更新(根を直接親にして参照距離を短く)
        size[x] = size[table[x]]
        return table[x]

def Unite(x,y):    #xとyを繋げる
    global size
    global rank
    x = Find(x)
    y = Find(y)
    sx = Size(x)
    sy = Size(y)

    if x == y:
        return

    if rank[x] > rank[y]:
        table[y] = x
        size[x] = sx + sy
    else:
        table[x] = y
        size[y] = sx + sy
        if rank[x] == rank[y]:
            rank[y] += 1

def Check(x,y):
    if Find(x) == Find(y):
        return True
    else:
        return False

def Size(x):
    return size[Find(x)]

N,M = inpl()

table = [i for i in range(N)]    #木の親 table[x] == x なら根
rank  = [1 for i in range(N)]    #木の長さ
size  = [1 for i in range(N)]    #集合のサイズ

for i in range(M):
    x,y,z = inpl()
    x -= 1
    y -= 1
    if not Check(x,y):
        Unite(x,y)

#print(table)

print(len(set(table)))