Problem: Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.

This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist). 

For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).

As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.

Help Miroslav turn over all n skewers.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.

Output

The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.

Examples

Input

7 2


Output

2
1 6 


Input

5 1


Output

2
1 4 

Note

In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.

In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations.

Solution (in JAVA):

Solution:
import java.io.BufferedInputStream;
import java.io.InputStream;
import java.util.ArrayList;


public class bb {
	
	public static void main(String[] args) {
		JoltyScanner in = new JoltyScanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		int best = Integer.MAX_VALUE / 2;
		ArrayList<Integer> ansar = new ArrayList<>();
		for (int i = 0; i <= k; i++) {
			int prev = i;
			int ans = 1;
			ArrayList<Integer> temp = new ArrayList<>();
			temp.add(i);
			for (int j = i + 1; j < n; j++) {
				if (j <= prev + k)
					continue;
				ans++;
				prev = j + k;
				if(prev >= n) ans = Integer.MAX_VALUE/2;
				temp.add(prev);
			}
			if (best > ans) {
				best = ans;
				ansar = temp;
			}
		}

		System.out.println(best);
		for (int v : ansar) {
			System.out.print((v + 1) + " ");

		}

	}

	static class JoltyScanner {
		public static final int BUFFER_SIZE = 1 << 16;
		public static final char NULL_CHAR = (char) -1;
		byte[] buffer = new byte[BUFFER_SIZE];
		boolean EOF_FLAG = false;
		int bufferIdx = 0, size = 0;
		char c = NULL_CHAR;
		BufferedInputStream in;

		public JoltyScanner(InputStream in) {
			this.in = new BufferedInputStream(in, BUFFER_SIZE);
		}

		public int nextInt() {
			long x = nextLong();
			if (x > Integer.MAX_VALUE || x < Integer.MIN_VALUE) {
				throw new ArithmeticException("Scanned value overflows integer");
			}
			return (int) x;
		}

		public long nextLong() {
			boolean negative = false;
			if (c == NULL_CHAR) {
				c = nextChar();
			}
			for (; !EOF_FLAG && (c < '0' || c > '9'); c = nextChar()) {
				if (c == '-') {
					negative = true;
				}
			}
			checkEOF();
			long res = 0;
			for (; c >= '0' && c <= '9'; c = nextChar()) {
				res = (res << 3) + (res << 1) + c - '0';
			}
			return negative ? -res : res;
		}

		public char nextChar() {
			if (EOF_FLAG) {
				return NULL_CHAR;
			}
			while (bufferIdx == size) {
				try {
					size = in.read(buffer);
					if (size == -1) {
						throw new Exception();
					}
				} catch (Exception e) {
					EOF_FLAG = true;
					return NULL_CHAR;
				}
				if (size == -1) {
					size = BUFFER_SIZE;
				}
				bufferIdx = 0;
			}
			return (char) buffer[bufferIdx++];
		}

		public void checkEOF() {
			if (EOF_FLAG) {
				throw new EndOfFileException();
			}
		}

		public class EndOfFileException extends RuntimeException {
		}
	}
}