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 math
n,k = map(int, input().split())
if k>n:
    k = n
if k == 0:
    print(n)
    for i in range(n):
        print(i+1,end=' ')
else:
    if n<=2*k+1:
        print(1)
        if n == k:
            print(n)
        else:
            print(n-k)
    else:
        print(int(math.ceil(n/(2*k+1))))
        l = [i for i in range(k+1,2*k+2)]
        d = (n%(2*k+1))
        if d == 0:
            d += 2*k+1
        if d<=k+1:
            e = l[0] - k
            print(e ,end=' ')
            v = int(math.ceil(n/(2*k+1))) - 2
            e += 2*k+1
            for i in range(v):
                print(e,end=' ')
                e += 2 * k + 1
            print(e)
        else:
            e = l[d-(k+1)] - k
            print(e ,end=' ')
            v = int(math.ceil(n/(2*k+1))) - 2
            e += 2*k+1
            for i in range(v):
                print(e,end=' ')
                e += 2 * k + 1
            print(e)