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 PYTHON):

Solution:
n, m = map(int, raw_input().split())
if 2 * m + 1 >= n:
  print 1
  print n / 2 + 1
else:
  dp = [10 ** 9] * (n + 5)
  way = [0] * (n + 5)
  for i in range(m + 1, n + 1):
    if i - 2 * m - 1 <= 0:
      dp[i] = 1
      way[i] = i - m
    else:
      dp[i] = dp[i - 2 * m - 1] + 1
      way[i] = i - m

  i = n - m - 1
  for j in range(m + 1):
    if n - j < 0:
      break
    if dp[n] > dp[i] + 1:
      dp[n] = dp[i] + 1
      way[n] = n - j
    i -= 1

  print dp[n]
  a = []
  i = n
  while i > 0:
    a.append(way[i])
    i = way[i] - m - 1

  print ' '.join(map(str, a[::-1]))