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 = map(int, input().split())
d = []
g = []
y = n
l = 0
for i in range(k + 1, k*2 + 1):
    d.append(i)
if k == 0:
    print(n)
    for i in range(n):
        print(i+1, end = ' ')
elif k >= n:
    print(1)
    print(n)
elif k*2 + 1 >= n:
    print(1)
    print(n // 2 + 1)
else:
    while n >= k*2 + 1:
        l += 1
        n -= k*2 + 1
        g.append(k*2 + 1)
    if n > 0 and n not in d:
        h = k*2 + 1
        while n not in d or h not in d:
            n += 1
            h -= 1
        print(l+1)
        
        print(h - k, end = ' ')
        p = h
        for i in range(l-1):
            print(p + k + 1, end = ' ')
            p += 2*k + 1
        print(y-(n- k - 1), end = ' ')
    elif n > 0 and n in d:
        print(l+1)
        l+=1
        print(n - k, end = ' ')
        p = n
        for i in range(l-1):
            print(p + k + 1, end = ' ')
            p = 2*k + 1  + p
    
    
    else:
        p = k+1
        print(l)
        for i in range(l):
            print(p, end = ' ')
            p += 2*k + 1