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
#pragma GCC optimize ("O3")
#pragma GCC optimize("unroll-loops")
#include"bits/stdc++.h"
using namespace std;
vector<bool> vis;
vector<vector<int>>g;
int n,m,u,v;
void dfs(int node)
{
	vis[node]=1;
	for(auto i:g[node])
	{
		if(!vis[i])
		{
			dfs(i);
		}
	}
}
int32_t main()
{
	
	cin>>n>>m;
	vis=vector<bool>(n);
	g=vector<vector<int>>(n);
	int z;
	for(int i=0;i<m;i++)
	{
		cin>>u>>v>>z;
		u--;v--;
		g[u].push_back(v);
		g[v].push_back(u);
		
	}
	int ans=0;
	for(int i=0;i<n;i++)
	{
		if(!vis[i]){ans++;dfs(i);}
	}
	
	cout<<ans;
	
	
	
}