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.util.Arrays;
import java.util.Scanner;
import java.util.stream.IntStream;

public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt(), m = scanner.nextInt();
			UnionFindTree uft = new UnionFindTree(n);
			IntStream.range(0, m).forEach(i -> {
				@SuppressWarnings("unused")
				int x = scanner.nextInt() - 1, y = scanner.nextInt() - 1, z = scanner.nextInt();
				uft.union(x, y);
			});
			System.out.println(IntStream.range(0, n).map(i -> uft.find(i)).distinct().count());
		}
	}

	/**
	 * Union Find Treeの実装
	 * 
	 * https://atcoder.jp/contests/abc040/submissions/4708654 を参考に作成
	 */
	private static class UnionFindTree {
		int[] parent;
		int[] count;

		UnionFindTree(int size) {
			parent = IntStream.range(0, size).toArray();
			count = new int[size];
			Arrays.fill(count, 1);
		}

		/**
		 * @param x
		 * @return xのroot
		 */
		int find(int x) {
			if (parent[x] == x) {
				return x;
			} else {
				int v = find(parent[x]);
				parent[x] = v;
				return v;
			}
		}

		/**
		 * @param x
		 * @param y
		 * @return xとyがつながっているかどうか
		 */
		@SuppressWarnings("unused")
		boolean same(int x, int y) {
			return find(x) == find(y);
		}

		/**
		 * xとyを結合する
		 * 
		 * @param x
		 * @param y
		 */
		void union(int x, int y) {
			int rootX = find(x);
			int rootY = find(y);
			if (rootX == rootY) {
				return;
			}
			parent[rootX] = rootY;
			parent[x] = rootY;
			count[rootY] += count[rootX];
		}

		/**
		 * @param x
		 * @return xと結ぶ都市の数
		 */
		@SuppressWarnings("unused")
		public int count(int x) {
			return count[find(x)];
		}
	}
}
