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;

struct unionFind{
	int n;vector<int>pr;
	unionFind(){};
	unionFind(int _n){
		n=_n;
		pr.resize(n,-1);
};
	int root(int a){
		if(pr[a]==-1)return a;
		else return pr[a]=root(pr[a]);
	}
	void unify(int a,int b){
		a=root(a);b=root(b);
		if(a==b)return;
		pr[a]=b;
	}	
	int connected(){return count(begin(pr),end(pr),-1);}
};

signed main(){
	int n,m;cin>>n>>m;
	unionFind uf(n);
	for(int i=0;i<m;i++){
		int x,y,z;
		cin>>x>>y>>z;x--;y--;
		uf.unify(x,y);
	}
	cout<<uf.connected()<<endl;
}