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:
# https://codeforces.com/problemset/problem/1040/B

def solution(N, K):
    solution = 0
    steps = []
    # 2K + 1 = skewer's maximum reach. For example if N = 5 and K = 2 then the max reach would be at skewer #3.
    if(N < 2 * K + 2):
        solution = 1
        if(N - K < 2):
            steps.append(1)
        else:
            steps.append(N // 2 + 1)
    else:
        if(N % (2 * K + 1) == 0):
            solution = int(N / (2 * K + 1))
            steps = list(range((1 + K), N+1, (2 * K + 1)))
        else:
            solution = N // (2 * K + 1) + 1
            if(N % (2 * K + 1) > (2 * K + 1) // 2):
                steps = list(range((1 + K), N+1, (2 * K + 1)))
            else:
                steps = list(range(1, N+1, (2 * K + 1)))                
            
    return solution, steps
    
N, K = [int(x) for x in input().split()]   
sol, steps = solution(N, K)
print(sol)
print(*steps)