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 <stdio.h>
#include <vector>
using namespace std;
int n, m;
int x, y, z;
int c;
vector<int> g[100000];
int s[100000];
void dfs(int a, int b) {
	s[a] = 1;
	for (int i = 0; i < g[a].size(); i++) {
		if (g[a][i] != b && s[g[a][i]] == 0)dfs(g[a][i], a);
	}
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++) {
		scanf("%d%d%d", &x, &y, &z);
		x--; y--;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	for (int i = 0; i < n; i++) {
		if (s[i] == 0) {
			c++;
			dfs(i, i);
		}
	}
	printf("%d\n", c);
}