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 = list(map(int, input().split()))

flip_n = 2*k + 1

rem = n % flip_n


order = []
l = 0
r = n
# arr = [1] * n

# def flip(arr, i):
# 	for j in range(max(0, i - k), min(n, i + k + 1)):
# 		arr[j] *= -1
# 	print(arr)

# print(arr)

# print(rem)
if flip_n >= n:
	order = [n // 2]
else:
	if rem != 0:
		if rem > k:
			# print('a')
			order.append(rem - k - 1)
			l = rem
			# flip(arr, rem - k - 1)
		else:
			# print('b')
			order.append(0)
			l = k + 1
			# flip(arr, 0)
			rem += k
			order.append(n - rem + k)
			r = n - rem + 1
			# flip(arr, n - rem + k)
	# print(l, r)
	for i in range(max(0, l + k), r, flip_n):
		# print(i)
		order.append(i)
		# flip(arr, i)

print(len(order))
print(*[i + 1 for i in order], sep=' ')
# print(l, r)