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:
#import resource
#import sys
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**26)
#sys.setrecursionlimit(0x1000000)
from sys import stdin, stdout
mod=(10**9)+7
mod1=mod-1
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):
    t=((fact[n])*(modinv(fact[r],p)%p)*(modinv(fact[n-r],p)%p))%p
    return t
def GCD(x, y):
   while(y):
       x, y = y, x % y
   return x
def BS(arr, l, r, x):
    if r >= l:
        mid = l + (r - l)/2
        if arr[mid] == x:
            return mid
        elif arr[mid] > x:
            return BS(arr, l, mid-1, x)
        else:
            return BS(arr, mid+1, r, x)
    else:n -1
from bisect import bisect_left as bl
from bisect import bisect_right as br
import itertools
import math
import heapq
from random import randint as rn
from Queue import Queue as Q
def comp(x,y):
    if(x[0]<y[0]):
        return -1
    elif(x[0]==y[0]):
        if(x[1]<y[1]):
            return -1
        else:
            return 1
    else:
        return 1
"""*********************************************************************************"""
n,k=map(int,raw_input().split())
z=(k*2)+1
w=n%z
if(w>=k+1):
    s=(n/z)+1
    q=[]
    t=k+1
    for i in range(s-1):
        q.append(str(t))
        t+=z
    r=n-(w-(k+1))
    q.append(str(r))
    print s
    print " ".join(q)
elif(w==0):
    print n/z
    for i in range(k+1,n+1,z):
        print i,
else:
    f=n
    q=[]
    s=0
    while(f>0):
        s+=1
        q.append(str(f))
        f-=z
    q=q[::-1]
    print s
    print " ".join(q)