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.

Examples

Input

3 1
1 2 1


Output

2


Input

6 5
1 2 1
2 3 2
1 3 3
4 5 4
5 6 5


Output

2


Input

100000 1
1 100000 100


Output

99999
from collections import deque

def solve():
    N, M = [int(_) for _ in raw_input().split()]
    to = [ [] for i in range(N) ]
    for i in range(M):
        Xi, Yi, Zi = [int(_) for _ in raw_input().split()]
        Xi -= 1
        Yi -= 1
        to[Xi].append(Yi)
        to[Yi].append(Xi)
    union = [None] * N
    que = deque()
    union_cnt = 0
    for i in range(N):
        if union[i] is not None:
            continue
        union_cnt += 1
        # BFS
        que.append(i)
        while len(que) > 0:
            vi = que.popleft()
            union[vi] = union_cnt
            for j in range(len(to[vi])):
                fri = to[vi][j]
                if union[fri] is not None:
                    continue
                union[fri] = union_cnt
                que.append(fri)
    print str(union_cnt)

if __name__ == '__main__':
    solve()
