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
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class Main {

	private static int[] root;

	public static void main(String[] args) {
		try {
			BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
			String[] input1 = bf.readLine().split(" ");
			int n = Integer.parseInt(input1[0]);
			int m = Integer.parseInt(input1[1]);

			// 計算式で結ばれているものは、Ax、Ayいずれか一つが分かればもう片方も分かる
			// 与えられた計算式からその親子関係をまとめる
			root = new int[n + 1];
			for (int i = 1; i <= m; i++) {
				String[] input2 = bf.readLine().split(" ");
				int ax = Integer.parseInt(input2[0]);
				int ay = Integer.parseInt(input2[1]);
				int axP = findMyRoot(ax);
				int ayP = findMyRoot(ay);

				if (root[ay] == 0) {

					// (? <- ... <-) Ax <- Ay
					if (axP != ay) {
						root[ay] = ax;
					}
				} else if (root[ax] == 0) {

					// (? <- ... <-) Ay <- Ax
					if (ayP != ax) {
						root[ax] = ay;
					}
				} else {

					// (? <- ... <-) Ax <- (? <- ... <-) Ay
					if (axP != ayP) {
						root[ayP] = ax;
					}
				}
			}

			// めくる必要がある回数を数える
			int cnt = 0;
			for (int i = 1; i <= n; i++) {
				if (root[i] == 0) {
					cnt++;
				}
			}
			System.out.println(cnt);
		} catch (IOException e1) {
		}
	}

	private static int findMyRoot(int child) {

		if (root[child] == 0) {
			return child;
		} else {
			return findMyRoot(root[child]);
		}
	}

}
