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
#include<bits/stdc++.h>
using namespace std;
const int MAX_N=1e5;
int N,M;
int f[MAX_N+1];
bool is_f[MAX_N+1];
int ans;
int find(int x){return f[x]?f[x]=find(f[x]):x;}
void merge(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
		f[fx]=fy;
	return;
}
int main()
{
	cin>>N>>M;
	for(int i=0;i<M;i++)
	{
		int X,Y,Z;
		cin>>X>>Y>>Z;
		merge(X,Y);
	}
	for(int i=1;i<=N;i++)
		is_f[find(i)]=true;
	for(int i=1;i<=N;i++)
		ans+=is_f[i];
    cout<<ans<<endl;
	return 0;
}