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 PYTHON):

Solution:
n,k = map(int,raw_input().split())
ile = 0
r = 0
pp = k+1
wyn = ''
wlicz = 0
if k == 0:
    print n
    for x in xrange(n+1):
        if x!=0:
            wyn += str(x)+' '
    print wyn
    exit(0)
if k*2+1 >= n:
    print 1
    print n/2+1
    exit(0)
else:
    ile = n/(k*2+1)
    #print 'ile = ',ile
    r = n%(k*2+1)
    #print 'r = ',r
    if r == 0:
        print ile
        for x in range(0, ile, +1):
            # print x
            wyn += str(pp) + ' '
            pp += k * 2 + 1
            wlicz += 1
        print wyn
        exit(0)
    while r<k+1:
        pp-=1
        r+=1
        if pp <=1:
            break
    for x in range(0,ile+1,+1):
        #print x
        wyn+=str(pp)+' '
        pp+=k*2+1
        wlicz+=1
print wlicz
print wyn