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:
#	!/bin/env python3
#	coding: UTF-8


#	✪ H4WK3yE乡
#	Mohd. Farhan Tahir
#	Indian Institute Of Information Technology and Management,Gwalior

#	Question Link
#	https://codeforces.com/problemset/problem/1040/B
#

# ///==========Libraries, Constants and Functions=============///


import sys

inf = float("inf")
mod = 1000000007


def get_array(): return list(map(int, sys.stdin.readline().split()))


def get_ints(): return map(int, sys.stdin.readline().split())


def input(): return sys.stdin.readline()

# ///==========MAIN=============///


def recurse(i, j):
    if i > n:
        if j >= n+1:
            return 0
        return inf
    ans = inf
    if dp[i][j] != -1:
        return dp[i][j]
    if i-j < k:
        d = min(ans, recurse(i+1, j))
        if d < ans:
            opti[i][j] = 0
            ans = d
    if j == 1 or i-j == k:
        d = min(ans, recurse(i+k+1, i+k+1)+1)
        if d < ans:
            opti[i][j] = 1
            ans = d
    dp[i][j] = ans
    return ans


if __name__ == "__main__":
    n, k = get_ints()
    dp = [[-1]*1005 for _ in range(1005)]
    opti = [[0]*1005 for _ in range(1005)]
    print(recurse(1, 1))
    i, j = 1, 1
    while i <= n:
        if opti[i][j] == 0:
            i += 1
        else:
            print(i, end=' ')
            j = i+k+1
            i = i+k+1