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,a,b=map(int,input().split())
s=[int(x) for x in input().split()]
temp=0
cst=0
for j in range(0,n//2):
    if(s[j]==2 and s[n-j-1]==2):
        cst=cst+(2*min(a,b))
    elif(s[j]==2 and s[n-j-1]==1):
        cst=cst+b
    elif(s[j]==1 and s[n-j-1]==2):
        cst=cst+b
    elif(s[j]==2 and s[n-j-1]==0):
        cst=cst+a
    elif(s[j]==0 and s[n-j-1]==2):
        cst=cst+a
    else:
        if(s[j]==1 and s[n-j-1]==0):
            temp=1
        if(s[j]==0 and s[n-j-1]==1):
            temp=1
if(temp==1):
    print('-1')
else:
    if(n%2==0):
        print(cst)
    else:
        if(s[n//2]==2):
            cst=cst+min(a,b)
        print(cst)"""

import math
n,k=map(int,input().split())
print(math.ceil(n/((2*k)+1)))
t=(2*k)+1
pos=k+1
rem=(n%t)
if(rem<(k+1) and rem!=0):
    pos=pos-(k+1-rem)
print(pos,end=" ")
while((pos+(2*k)+1)<=n):
    pos=(pos+(2*k)+1)
    print(pos,end=" ")
print(" ")


"""if(n<=t):
    
print(pos,end=" ")
while((pos+(2*k)+1)<=n):
    pos=(pos+(2*k)+1)
    print(pos,end=" ")
print(" ")"""