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:
import sys
# from collections import deque
# from collections import Counter
# from math import sqrt
# from math import log
from math import ceil
# from bisect import bisect_left, bisect_right

alpha=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
# mod=10**9+7
# mod=998244353

# def BinarySearch(a,x): 
# 	i=bisect_left(a,x) 
# 	if(i!=len(a) and a[i]==x): 
# 		return i 
# 	else: 
# 		return -1

# def sieve(n): 
# 	prime=[True for i in range(n+1)]
# 	p=2
# 	while(p*p<=n): 
# 		if (prime[p]==True): 
# 			for i in range(p*p,n+1,p): 
# 				prime[i]=False
# 		p+=1
# 	prime[0]=False
# 	prime[1]=False
# 	s=set()
# 	for i in range(len(prime)):
# 		if(prime[i]):
# 		s.add(i)
# 	return s

# def gcd(a, b):
# 	if(a==0):
# 		return b 
# 	return gcd(b%a,a)

fast_reader=sys.stdin.readline
fast_writer=sys.stdout.write

def input():
	return fast_reader().strip()

def print(*argv):
	fast_writer(' '.join((str(i)) for i in argv))
	fast_writer('\n')

#____________________________________________________________________________________________________________________________________

def find(s1,s2):
	for i in alpha:
		if(i!=s1 and i!=s2):
			return i

for _ in range(1):
	n,k=map(int, input().split())
	ans=ceil(n/(2*k+1))
	print(ans)
	if(n%(2*k+1)<=k+1 and n%(2*k+1)>0):
		ans2=n%(2*k+1)
	else:
		ans2=k+1
	l=[]
	while(ans2<=n):
		l.append(ans2)
		ans2+=2*k+1
	print(*l)