Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

3 1
1 2 1

Sample Output 1

2

You can determine all of A_1, A_2, A_3 by using the magic for the first and third cards.

Sample Input 2

6 5
1 2 1
2 3 2
1 3 3
4 5 4
5 6 5

Sample Output 2

2

Sample Input 3

100000 1
1 100000 100

Sample Output 3

99999

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define M 1000000007
#define F first
#define S second
#define mp make_pair
#define pb push_back
#define gcd __gcd
#define in(a) scanf("%d",&a)
#define in2(a,b) scanf("%d%d",&a,&b)
#define in3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define read(v,n) for(h=0;h<n;h++)in(v[h]);
#define sc(ch) scanf("%c",&ch)
#define sstr(str) scanf("%s",str)
#define pr(n) printf("%d ",n)
#define out(n) printf("%d\n",n)
#define cf(n) printf("%I64d\n",n)
#define yes printf("YES\n")
#define no printf("NO\n")
#define lin printf("\n")
#define dbg(v,n) for(h=0;h<n;h++)pr(v[h]); lin
#define ck printf("continue\n")
#define ii pair<int,int>
#define vi vector<int>
#define vii vector<pair<int,int> >
#define vvi vector<vector<int> >
#define viii vector<pair<pair<int,int>,int> >
#define vvii vector<vector<pair<int,int> > >
#define all(vec) vec.begin(),vec.end()
#define asc(vec) sort(vec.begin(),vec.end())
#define lp(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define lower(v,k) lower_bound(v.begin(),v.end(),k)-v.begin()
#define upper(v,k) upper_bound(v.begin(),v.end(),k)-v.begin()
#define N 100005
typedef long long int ll;
int par[N],lev[N],tot[N];
void makeset(int n)
{
    fill(par,par+n,-1);
    fill(tot,tot+n,1);
    return;
}
int findset(int x)
{
    int y=x;
    while(par[y]>=0)
        y=par[y];
    if(x!=y)
        par[x]=y;
    return y;
}
void combine(int a,int b)
{
    a=findset(a);
    b=findset(b);
    if(a!=b)
    {
        if(lev[a]<lev[b])
            par[a]=b,tot[b]+=tot[a];
        else if(lev[a]>lev[b])
            par[b]=a,tot[a]+=tot[b];
        else
            lev[b]++,par[a]=b,tot[b]+=tot[a];
    }
    return ;
}
int main()
{
    int i,n,m,a,b,c;
    in2(n,m);
    makeset(n);
    while(m--)
    {
        in3(a,b,c);
        combine(a-1,b-1);
    }
    a=0;
    for(i=0;i<n;i++)
        if(par[i]<0)a++;
    out(a);
}