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(x) for x in input().split()]

if k==0:
	print(n)
	for i in range(n):
		print(i+1,end=" ")
	exit()
marks = []
flag = 0
a = []
if n//2<=k:
	marks.append(n//2+1)
	flag = 1
else:

	for i in range(k+1):
		o = [0]*n
		mark = []
		j = i
		while j<n:
			mark.append(j)
			for p in range(j,j+k+1):
				if p<n:
					o[p] = 1
			for p in range(j-k-1,j):
				if p>=0:
					o[p] = 1
			j = j + 2*k + 1
		try:
			#print(o,i+1)
			x = o.index(0)
		except:
			a.append((len(mark),len(a),mark))
			#print(o,mark)
if flag == 1:
	print(1)
	print(marks[0])
else:
	#print(a)
	ss = min(a)
	#print(ss)
	mark = ss[2]
	print(len(mark))
	for i in range(len(mark)):
		print(mark[i]+1,end=" ")