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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include<limits.h>
#include<stdio.h>
#include<cstring>
#include<string>
using namespace  std;
typedef long long ll;
typedef long double ld;
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pb push_back
#define mp make_pair
#define sz(a) a.size()
#define fi first
#define se second
#define rsz(a,n) a.resize(n)
#define test(t) ll t;cin>>t;while(t--)
#define forn(i,e) for(int i = 0; i < e; i++)
#define forsn(i,s,e) for(int i = s; i < e; i++)
#define rforn(i,s) for(int i = s; i >= 0; i--)
#define rforsn(i,s,e) for(int i = s; i >= e; i--)
#define fillv(a,k) memset(a,k,sizeof(a))
#define leadzero(a) __builtin_clz(a) //count leading zeros
#define trailzero(a) __builtin_ctz(a) //count trailing zeros
#define bitcount(a) __builtin_popcount(a) // count set bits (add ll)
#define ln cout<<"\n"
// #define sp cout<<" "
#define spaceprint(a,i,s,n) {forsn(i,s,n) { cout<<a[i]; sp;}}
#define lineprint(a,i,s,n) {forsn(i,s,n) {cout<<a[i]; ln;}}
#define maxe(a) *max_element(a.begin(),a.end())
#define maxi(a) max_element(a.begin(),a.end())-a.begin()
#define mine(a) *min_element(a.begin(),a.end())
#define mini(a) min_element(a.begin(),a.end())-a.begin()
#define all(c) c.begin(),c.end()
#define trav(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define present(container, element) (container.find(element) != container.end())
#define cpresent(container, element)  (find(all(container),element) != container.end())// check the presence of element
//copy(from_begin, from_end, to_begin); copy function
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<ll,ll> p64;
typedef pair<int,int> p32;
typedef pair<int,p32> p96;
typedef vector<ll> v64;
typedef vector<string> vs;
typedef vector<int> v32;
typedef vector<v32> vv32;
typedef vector<v64> vv64;
typedef vector<p32> vp32;
typedef vector<p64> vp64;
typedef vector<vp32> vvp32;
typedef map<int,int> m32;
typedef map<ll,ll> m64;
const int LIM = 1e5+5, MOD =  1e9;
#define sumv(v) accumulate(all(v),ll(0))
#define productv(v) accumulate(all(v), ll(1), multiplies< ll >())
ll gcd(ll a, ll b) { if(b == 0) return a; return gcd(b, a % b); }
ll fastpowMOD(ll a, ll p,ll MOD){ if(p==0) return 1; ll z = fastpowMOD(a,p/2,MOD); z = (z*z)%MOD; if(p%2) z = (z*a)%MOD; return z; }


bool compare (string s1,string s2)
{
	return s1.length()<s2.length();
}

ll a[100004];
ll size[100004];

ll root(ll x)
{
	while (a[x]!=x)
		x=a[x];
	return x;
}

ll join (ll x,ll y)
{
	if (size[root(x)]<size[root(y)])
	{
		a[root(x)]=root(y);
		size[root(y)]+=size[root(x)];
	}
	else
	{
		a[root(y)]=root(x);
		size[root(x)]+=size[root(y)];	
	}
}

int main()
{
	fast;
	ll n,m;
	cin>>n>>m;
	for (ll i=1;i<=n;i++)
	{
		a[i]=i;
		size[i]=1;
	}
	for (ll i=0;i<m;i++)
	{
		ll x,y,z;
		cin>>x>>y>>z;
		if (root(x)!=root(y))
			join(x,y);
	}
	set <ll> s;
	for (ll i=1;i<=n;i++)
		s.insert(root(i));
	cout<<s.size();
	

}