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.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution 
{
	static class InputReader
	{
	    private final InputStream stream;
	    private final byte[] buf = new byte[8192];
	    private int curChar, snumChars;

	    public InputReader(InputStream st) {
	      this.stream = st;
	    }

	    public int read() {
	      if (snumChars == -1)
	        throw new InputMismatchException();
	      if (curChar >= snumChars) {
	        curChar = 0;
	        try {
	          snumChars = stream.read(buf);
	        } catch (IOException e) {
	          throw new InputMismatchException();
	        }
	        if (snumChars <= 0)
	          return -1;
	      }
	      return buf[curChar++];
	    }

	    public int nextInt() {
	      int c = read();
	      while (isSpaceChar(c)) {
	        c = read();
	      }
	      int sgn = 1;
	      if (c == '-') {
	        sgn = -1;
	        c = read();
	      }
	      int res = 0;
	      do {
	        res *= 10;
	        res += c - '0';
	        c = read();
	      } while (!isSpaceChar(c));
	      return res * sgn;
	    }

	    public long nextLong() {
	      int c = read();
	      while (isSpaceChar(c)) {
	        c = read();
	      }
	      int sgn = 1;
	      if (c == '-') {
	        sgn = -1;
	        c = read();
	      }
	      long res = 0;
	      do {
	        res *= 10;
	        res += c - '0';
	        c = read();
	      } while (!isSpaceChar(c));
	      return res * sgn;
	    }

	    public int[] nextIntArray(int n) {
	      int a[] = new int[n];
	      for (int i = 0; i < n; i++) {
	        a[i] = nextInt();
	      }
	      return a;
	    }

	    public String readString() {
	      int c = read();
	      while (isSpaceChar(c)) {
	        c = read();
	      }
	      StringBuilder res = new StringBuilder();
	      do {
	        res.appendCodePoint(c);
	        c = read();
	      } while (!isSpaceChar(c));
	      return res.toString();
	    }

	    public String nextLine() {
	      int c = read();
	      while (isSpaceChar(c))
	        c = read();
	      StringBuilder res = new StringBuilder();
	      do {
	        res.appendCodePoint(c);
	        c = read();
	      } while (!isEndOfLine(c));
	      return res.toString();
	    }

	    public boolean isSpaceChar(int c) {
	      return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	    }

	    private boolean isEndOfLine(int c) {
	      return c == '\n' || c == '\r' || c == -1;
	    }
	}
	public static void main(String[] args) throws Exception
	{
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int jump = 1 + 2*k;
        if(jump > n || k > n)
        {
                System.out.println("1");
        	System.out.println(n/2 + 1);
        	return;
        }
        ArrayList<Integer> ans = new ArrayList<>();
        int min = Integer.MAX_VALUE;
        for(int i=1;i<=k+1;i++)
        {
            int count = 1;
            ArrayList<Integer> arr = new ArrayList<>();
            arr.add(i);
            
            for(int j=i+2*k+1;j<=n;j+=1+2*k)
            {
                arr.add(j);
                count++;
            }
            if(count < min && n - arr.get(arr.size() - 1) <= k)
            {
                min = count;
                ans = new ArrayList<Integer>(arr);
            }
        }
        //ArrayList<Integer> ans = new ArrayList<>();
        System.out.println(ans.size());
        for(int g=0;g<ans.size();g++)
            System.out.print(ans.get(g) + " ");
        System.out.println();
     }
}