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,m=map(int, input().split())
k=(m+1)-1
v=n
m1=m+1
st=''
p=False
c=m+m+1
q=n//c
if n%c!=0:
    q+=1
ar=[]
w=n-m-1
r=w//c
print (q)


if n<=m:
    print ((n+1)//2)
elif m==0:
    for i in range(1,n+1):
        st+=str(i)+' '
    print(st)
else:
    while v%c!=0:
        k+=1
        p=True
        v=n-k
    
    if p:
       if k>c:
            g=k-m-1
            h=m+1
            st+=str(g-m)+' '
            for i in range (g,n):
                if (i-g)%c==(c+1)//2:
                    st+=str(i)+' '
            st+=str(n)
            print(st)
       else:
            g=k
            st+=str(k-m)+' '
            for i in range (k+1,n+1):
                if (i-k)%c==(c+1)//2:
                    st+=str(i)+' '
            print(st)
    else:
        for i in range(1,n+1):
            if i%c==(c+1)//2:
                st+=str(i)+' '
        print (st)