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:
import array
n,k = map(int,input().split())
kt= array.array('i',(int(0) for i in range(0,2010)))

kt[0] = 1
for i in range(k+1,k*2+2):
    kt[i] = 1
fs,mid,en =0,0,0
div = k*2+1
if k == 0:
    print(n)
    for i in range(1,n+1):
        print(i,end=' ')
elif k >= n/2:
    print(1)
    print(int(n/2+1))
else:
    for i in range(k+1,k*2+2):
        #print(kt[i],' ',i)
        if kt[(n-i)%div] == 1:
            fs = i
            en = (n-i)%div
            mid = n-fs-en
            sum = int(0)
            if(fs > 0):
                sum+=1
            if(en > 0):
                sum+=1
            if(mid > 0):
                sum += int(mid/div)
            else:
                sum += 0
            print(sum)
            #First
            print(fs-k,end =' ')
            cnt = fs
            while(cnt+k+1 <= n):
                print(cnt+k+1,end=' ')
                if(mid>0):
                    cnt+=div
                else:
                    cnt+=en
            break