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

题意

陈师傅在烤羊肉串，为了烤的更加均匀，他每隔一段时间就要翻一下，而他每次可以翻的范围为【i-k，i+k】，题目中给定陈师傅有n串，写一个程序帮助陈师傅计算最少几次可以将所有的串翻一遍，并把每次翻的点记录下来。。。。。（终于，忙碌了两个小时的陈师傅打开了美团外卖·······）

思路

显然，每次可以翻2k+1个，所以若n<2k+1，只需一次就可以；如果k=0，则需要n次；然后就通过对其取模判断，若n%（2*k+1）大于k或等于0，则从k+1开始翻，若小于等于k则从第一个开始翻。`

#include<cstdio>
#define maxn 1010
using namespace std;

int a[maxn];
	int n,k,cnt;
int main()
{
	scanf("%d%d",&n,&k);
	if(!k){
		cnt=n;
		for(int i=1;i<=n;i++)a[i]=i;
	}
	else if(n<=2*k+1){
		cnt=1;
		a[1]=n/2+1;
	}
	else if(n>2*k+1){
		if(n%(2*k+1)>k||!(n%(2*k+1))){
			cnt=0;
			for(int i=k+1;i<=n;i+=(2*k+1))a[++cnt]=i;
		}
		else {
			cnt=0;
			for(int i=1;i<=n;i+=(2*k+1))a[++cnt]=i;
		}
	}
	printf("%d\n",cnt);
	for(int i=1;i<=cnt;i++)printf("%d ",a[i]);
	return 0;
}