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 <iostream>
#include <algorithm>
#include <numeric>
using namespace std;
int a[100000], x[100000], y[100000], z[100000], n, m, cnt;

int f(int x) {
  if(x == a[x])
    return x;
  return a[x] = f(a[x]);
}

void con(int p, int t) {
  if(f(p) == f(t))
    return;
  --cnt;
  a[f(p)] = f(t);
}

int main() {
  cin >> n >> m;
  cnt = n;
  for(int i = 0; i < m; ++i)
    cin >> x[i] >> y[i] >> z[i], (--x[i], --y[i]);
  iota(a, a + n, 0);
  for(int i = 0; i < m; ++i)
    con(x[i], y[i]);
  cout << cnt << endl;
}