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:
R = lambda :list(map(int,input().split()))
n,k = R()
t = (n//(2*k+1))
if n%(2*k+1) == 0 :
	l=[(k+1 + (2*k+1)*i) for i in range(0,t)]
	print(t)
	print(*l ,sep =' ')
else :
	r = n - t*(2*k + 1)
	if r >= k + 1:
		l = [(k+1 + (2*k+1)*i) for i in range(0,t)]
		l.append(n-(r-k-1))
	else :
		p = (n-(k+1))//(2*k+1)
		if (n-(k+1))%(2*k+1) == 0 :
			l = [(1+ (2*k+1)*i) for i in range(0,p+1)]
		else :
			l =[(1+ (2*k+1)*i) for i in range(0,p+1)]
			r = (n-(k+1)) - p*(2*k+1)
			l.append(n-(r-k-1))

	print(t+1)
	print(*l,sep = ' ')