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 ll long long
vector<ll> adj[200007] ;
ll vis[200007] ;

void dfs(ll n ) {
    vis[n]=1 ;
    for (auto it : adj[n]) {
        if (!vis[it]) dfs(it) ;
    }

}

int main() {

    ll n , m ;
    cin >> n>> m ;
    while(m--) {
        ll a , b ,c ;
        cin >> a >> b >> c ;
        adj[a].push_back(b) ;
        adj[b].push_back(a) ;
    }
    ll ans = 0 ;
    for (ll i =1 ; i<=n ;i++) {
        if (!vis[i]) {
            dfs(i) ;
            ans++ ;
        }
    }

    cout << ans << endl ;

    return 0 ;
}
