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
#coding:utf-8
# https://reiyw.com/post/union-find-tree-in-python/
import collections
import itertools
import operator
class UnionFind:
    def __init__(self, elems=None):
        class KeyDict(dict):
            def __missing__(self, key):
                self[key] = key
                return key

        self.parent = KeyDict()
        self.rank = collections.defaultdict(int)

        if elems is not None:
            for elem in elems:
                _, _ = self.parent[elem], self.rank[elem]

    def find(self, x):
        if self.parent[x] == x:
            return x
        else:
            self.parent[x] = self.find(self.parent[x])
            return self.parent[x]

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if self.rank[x] < self.rank[y]:
            self.parent[x] = y
        else:
            self.parent[y] = x
        if self.rank[x] == self.rank[y]:
            self.rank[x] += 1

    def are_same(self, x, y):
        return self.find(x) == self.find(y)

    def grouper(self):
        roots = [(x, self.find(x_par)) for x, x_par in self.parent.items()]
        root = operator.itemgetter(1)
        for _, group in itertools.groupby(sorted(roots, key=root), root):
            yield [x for x, _ in group]

N,M=map(int,raw_input().split())
uf=UnionFind(range(1,N+1))
for i in range(M):
    x,y,z=map(int,raw_input().split())
    uf.unite(x,y)

from collections import defaultdict
F=set()
for i in range(1,N+1):
    F.add(uf.find(i))
print(len(F))    