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:
from math import ceil

if __name__ == '__main__':
    n, k = [int(__) for __ in input().strip().split()]
    if k == 0:
        print(n)
        for i in range(n):
            print(i + 1, end=" ")
    else:
        if n <= 2 * k + 1:
            print(1)
            print(ceil(n / 2))
        else:
            l = []
            co = 0
            arr = [i + 1 for i in range(n)]
            while len(arr) > 0:
                if len(arr) > 2 * k + 1:
                    co += 1
                    l.append(arr[k])
                elif len(arr) == 2 * k + 1:
                    co += 1
                    l.append(arr[k])
                elif len(arr) <= 2 * k + 1:
                    break
                arr = arr[2 * k + 1:]
            if len(arr) > 0:
                if len(arr) <= k:
                    co += 1
                    for i in range(len(l)):
                        l[i] -= (k - len(arr) + 1)
                    l.append(arr[-1])
                else:
                    co += 1
                    l.append(arr[k])
            print(co)
            print(' '.join([str(i) for i in l]))