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:
from fractions import gcd
from math import factorial, ceil, sqrt, atan2, log, pi, e, asin, acos, cos, sin, floor
from itertools import *
from fractions import Fraction
import string
import copy
import random
import bisect
from decimal import *
from collections import deque
from sys import *
digs = string.digits + string.ascii_letters

def id_generator(size=20, chars=string.digits):
    return ''.join(random.choice(chars) for _ in range(size))
 
def mp():
    return map(int,str(raw_input()).split())



n,k=mp()
if k>=n/2:
	print 1
	print (n+1)/2
else:
	swap=[0 for i in range(n)]
	x=n%(2*k+1)
	y=ceil(n/(2.0*k+1))
	print int(y)

	a=x
	b=(x+(2*k+1))-(k+1)

	i=0
	ans=0
	val=0
	#print a,b,x
	while (a+(2*k+1)*(y-1)+k)<n:
		a+=1
		#print a
	while a>k+1:
		a-=1
	while i<y:
		if i==0:
			val+=a
			print val,
		else:
			val+=2*k+1
			print val,
		i+=1