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()]

'''
1 2 3 4 5 6 7
1 2 3 | 4 5 6 7
split into 2k groups

if last group < k then problem

if k = 3, n = 7

1 2 3 4 5 6 7

1 2 3 4 5

first sz varies from k+1 to 2*k+1


1 (k+1 - 2k+1) + x(2*k) + 1(2k+1 - k+1)
1 2 3 4 5 6 7 8 9 10

1 2 3 4 5
11,2 = 1 6 10
8, 2 = 1 6
'''

places, placee = 0, 0
ans = 2*n
for start in range(k+1, 2*k+2):
    for end in range(k+1, 2*k+2):
        if (n - (start) - (end))%(2*k+1) == 0:
            #print(n,start,end,k)
            if 2 + (n-(start)-(end))//(2*k+1) < ans:
                ans = min(ans, 2 + (n-(start)-(end))//(2*k+1))
                places = start
                placee = end

print(ans)
print(places-k, end=' ')
if ans>1:
    for j in range((n-places-placee)//(2*k+1)):
        print(places-k + (j+1)*(2*k+1),end=' ')
    last = placee - k -1 
    print(n-last)
    #print(places - k + (j+1)*(2*k+1))