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;
const int maxn = 300001;
int m, n, ans;
bool visited[maxn];
vector < int > edge[maxn];
void dfs(int now){
	if(visited[now])return;
	visited[now] = true;
	for(auto &i : edge[now])dfs(i);
}
int main(){
	cin >> n >> m;
	for(int a, b, c, i = 0;i < m;++i){
		cin >> a >> b >> c;
		edge[a].push_back(b);
		edge[b].push_back(a);
	}
	for(int i = 1;i <= n;++i){
		if(!visited[i]){
			++ans;
			dfs(i);
		}
	}
	cout << ans << '\n';
}
