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;
#define rep(i,N) for(int i=0; i<(N); i++)
#define pb push_back
typedef long long ll;
const ll mod=1e9+7;

ll N,M;
ll A[100005];
ll mada[100005];
ll ans;
vector<vector<ll> > vv(100005);

void dfs(ll x){
	if(mada[x]) return;
	mada[x]=1;
	rep(i,vv[x].size()){
		ll y=vv[x][i];
		dfs(y);
	}
	return;
}

int main(){ 

cin.tie(0);
ios::sync_with_stdio(false);

cin>>N>>M;

rep(i,M){
	ll x,y,z;
	cin>>x>>y>>z;
	x--; y--;
	vv[x].pb(y);
	vv[y].pb(x);
}

rep(i,N){
	if(mada[i]==0){
		ans++;
		dfs(i);
	}
}

cout<<ans<<endl;

}
