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: Time Limit Exceeded)

list=input().split(" ")
n=int(list[0])
m=int(list[1])

siki=[]

for i in range(m):
    list=input().split(" ")
    a=int(list[0])
    b=int(list[1])
    siki.append([a,b])

#print(siki)
#siki.sort(key=itemgetter(1))
#print(siki)

katamari=[]
katamari.append(siki[0])
dekai=siki[0][1]
nanko=1

siki=siki[1:]

for sk in siki:
    aswitch=0
    bswitch=0
    ais=0
    bis=0
    #print(katamari)
    #print(nanko)
    for kt in katamari:
        if sk[0] in kt:
            aswitch=1
            break
        else:
            ais+=1
    for kt in katamari:
        if sk[1] in kt:
            bswitch=1
            break
        else:
            bis+=1
    if aswitch==0:
        if bswitch==0:
            katamari.append(sk)
            nanko+=1
        else:
            katamari[bis].append(sk[0])
            nanko+=1
    else:
        if bswitch==0:
            katamari[ais].append(sk[1])
            nanko+=1
        else:
            if ais!=bis:
                sho=min(ais, bis)
                dai=max(ais, bis)
                for kd in katamari[dai]:
                    katamari[sho].append(kd)
                if sho+1>=len(katamari):
                    katamari=katamari[:dai]
                    nanko+=1
                else:
                    katamari=katamari[:dai]+katamari[dai+1:]
                    nanko+=1
print(n-nanko)