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.InputStreamReader;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		int m = Integer.parseInt(sa[1]);

		UnionFind uf = new UnionFind(n);
		for (int i = 0; i < m; i++) {
			sa = br.readLine().split(" ");
			int x = Integer.parseInt(sa[0]) - 1;
			int y = Integer.parseInt(sa[1]) - 1;
			uf.union(x, y);
		}
		br.close();

		System.out.println(uf.num);
	}

	static class UnionFind {
		int[] parent, size;
		int num = 0; // 連結成分の数

		UnionFind(int n) {
			parent = new int[n];
			size = new int[n];
			num = n;
			for (int i = 0; i < n; i++) {
				parent[i] = i;
				size[i] = 1;
			}
		}

		void union(int x, int y) {
			int px = find(x);
			int py = find(y);
			if (px != py) {
				parent[px] = py;
				size[py] += size[px];
				num--;
			}
		}

		int find(int x) {
			if (parent[x] == x) {
				return x;
			}
			parent[x] = find(parent[x]);
			return parent[x];
		}

		/**
		 * xとyが同一連結成分か
		 */
		boolean same(int x, int y) {
			return find(x) == find(y);
		}

		/**
		 * xを含む連結成分のサイズ
		 */
		int size(int x) {
			return size[find(x)];
		}
	}
}
