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()))
max = k*2+1
min = k+1
if k == 0:
  print(n)
  for x in range(1, n):
    print(x ,end=' ')
  print(n)
elif n % max == 0:
  print(int(n / max))
  for x in range(max - k, n+1, max):
    if(x + max >= n):
      print(x)
    else:
      print(x, end=' ')
elif k >= n:
  print(1)
  print(1)
else:
  for i in range(n-min,-1, -1):
    if i % max == 0:
      c = i/max
      r = n - i
      if r >= min and r <= max:
        print(int(c+1))
        for x in range(r - k, n, max):
          if(x + max >= n):
            print(x)
          else:
            print(x, end=' ')
      else:
        for rem in range(0, r+1):
          if(rem >= min and rem <= max and r-rem >= min and r-rem <= max):
            print(int(c+2))
            for x in range(rem - k, n+1, max):
              if(x + max >= n+1):
                print(x)
              else:
                print(x, end=' ')
            break
      break