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:
n,k =  [int(i) for i in input().split(' ')]

if k >= n//2:
	print(1)
	print(max(n//2 + 1, 1))
	exit(0)

mark = [(i+1) for i in range(n)]

for i in range(2*k + 1):
	if (n + i) % ((2*k)+1) == 0:
		n += i
		break

if i > k:
	mark = [-1 for i in range(k)] + mark + [-1 for i in range(i - k)]
elif i == k:
	[-1 for i in range(k)] + mark
elif i == 0:
	pass
else:
	[-1 for i in range(i)] + mark


idx = [False for i in range(len(mark))]

prev = 1
for i in range(1, (n // ((2*k)+1)) + 1):
	high = i * ((2*k)+1) 

	idx[(high+prev)//2 - 1] = True

	prev = high+1


print(sum(idx))
for i in range(len(idx)):
	if idx[i] is True and mark[i] != -1:
		print(mark[i], end = ' ')