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:
num_input = input()
num_input = num_input.split(" ")

n = int(num_input[0])
k = int(num_input[1])

res = 0
res_temp = []
min = k + 1
max = 2*k + 1


if n <= max:
    print(1)
    print(int(n / 2) + 1)
else:
    if n % max == 0:
        res = int(n / max)
        for i in range(0, res):
            res_temp.append(max)
    else:
        res = int(n / max) + 1
        
        if n % max < min:
            for i in range(0, res - 2):
                res_temp.append(max)
            
            temp = n % max + max
            res_temp.append(min)
            res_temp.append(temp - min)
        else:
            for i in range(0, res - 1):
                res_temp.append(max)
            res_temp.append(n % max)


    res_temp.insert(0,res_temp.pop())

    print(res)
    #print(res_temp)

    dp = []
    dp.append(1 + res_temp[0] - min)
    
    for i in range(1, len(res_temp)):
        dp.append(dp[i - 1] + max)
    
    for i in range(0, len(dp)):
        dp[i] = str(dp[i])

    #print(dp)
    ouput = " ".join(dp)
    print(ouput)