CF1040B Shashlik Cooking

原题链接：http://codeforces.com/contest/1040/problem/B

Shashlik Cooking

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 $i-k,i-k+1,\cdots ,i-1,i+1,\cdots ,i+k-1,i+k$ (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 $k(1\le n\le 1000,0\le k\le 1000)$ — 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.

题解

直接对$2\times k+1$取模，如果余数大于$k$，就可以直接在后面补一个；否则我们就把第一个也往前调一点。

代码

#include<bits/stdc++.h>
using namespace std;
const int M=1e3+5;
int n,k,st=1;
void in(){scanf("%d%d",&n,&k);}
void ac()
{
	if(n%(2*k+1)>k)st=k+1,printf("%d\n",n/(2*k+1)+1);
	else if(n%(2*k+1)==0)st=k+1,printf("%d\n",n/(2*k+1));
	else printf("%d\n",n/(2*k+1)+1);
	for(;st<=n;st+=2*k+1)printf("%d ",st);
}
int main(){in(),ac();}