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;
using ll=long long;
#define rep(i,n) for(ll i=0;i<n;i++)
const ll INF=1e15;
const int inf=1e9;
const ll MOD=1e9+7;

int par[100005];
int root(int i){
    if(par[i]==i)return i;
    else return par[i]=root(par[i]);
}
int main(){
    int N,M;cin>>N>>M;
    rep(i,N)par[i]=i;
    rep(i,M){
        int a,b,c;cin>>a>>b>>c;
        a--;b--;
        a=root(a);b=root(b);
        if(a!=b)par[par[b]]=par[a];
    }
    rep(i,N)par[i]=root(i);
    set<int> ans;
    rep(i,N)ans.insert(par[i]);
    cout<<ans.size()<<endl;
    return 0;
}