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 PYTHON3):

Solution:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sat Dec 21 18:17:38 2019

@author: puchisspanitpotjaman
"""
data = input()
a = [int(n) for n in data.split()]
opt = 0
opt_n = 0
opt_k = 0
opt_l = 0
if a[0] <= 2*a[1] + 1:
	print(1)
	print(a[0]//2+1)
elif a[0] < 4*a[1] + 2 and a[0] > 2*a[1] + 1:
	print(2)
	if a[0]%2 == 1:
		opt = a[0]//2
		opt_n = opt - a[1]
		opt_k = opt + 1
		opt_l = opt_k - a[1]
		print(opt_n,a[0]-opt_l+1)
	if a[0]%2 == 0:
		opt = a[0]//2
		opt_n = opt - a[1]
		print(opt_n,a[0]-opt_n+1)
else:
	i = a[0]
	col = 0
	while i > 4*a[1]+2:
		i -= 2*a[1]+1
		col += 1
	print(col+2)
	if i%2 == 1:
		opt = i//2
		opt_n = opt - a[1]
		opt_k = opt + 1
		opt_l = opt_k - a[1]
		print(opt_n, end =" ")
		l = opt_n
		for k in range(col):
			l += 2*a[1]+1
			print(l, end =" ")
		print(a[0]-opt_l+1)
	
	else:
		opt = i//2
		opt_n = opt - a[1]
		l = opt_n
		print(opt_n, end =" ")
		l = opt_n
		for k in range(col):
			l += 2*a[1]+1
			print(l, end =" ")
		print(a[0]-opt_n+1)