CF1040B.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 $i - k$, $i - k + 1$, ..., $i - 1$, $i + 1$, ..., $i + k - 1$, $i + k$ (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.

长话短说，沙希利克（shashlik）是米罗斯拉夫最喜爱的食物。沙希利克是同时在若干串签上烤制的。每根串签有两种状态：初始状态和翻转状态。

这一次，米罗斯拉夫将 $n$ 根串签彼此平行地排成一排，并从左到右依次用连续整数 $1$ 到 $n$ 对其编号。为了使烤制效果更佳，他将串签摆放得非常靠近，因此当他翻转第 $i$ 根串签时，会导致其左右两侧各自最近的 $k$ 根串签也被翻转，即串签编号为 $i - k$, $i - k + 1$, ..., $i - 1$, $i + 1$, ..., $i + k - 1$, $i + k$ 的那些串签（如果存在的话）。

例如，设 $n = 6$ 且 $k = 1$。当米罗斯拉夫翻转第 $3$ 根串签时，则编号为 $2$、$3$、$4$ 的串签都会被翻转。若此后他再翻转第 $1$ 根串签，则编号为 $1$、$3$、$4$ 的串签将被翻转，而编号为 $2$ 的串签将恢复至初始状态（因为它又被翻转了一次）。

如前所述，烹饪的艺术要求精准的时机控制，因此米罗斯拉夫希望以最少的操作次数翻转全部 $n$ 根串签。例如，在上述例子中（$n = 6$ 且 $k = 1$），两次翻转操作即足够：他可以翻转第 $2$ 根和第 $5$ 根串签。

请帮助米罗斯拉夫翻转全部 $n$ 根串签。

The first line contains two integers $n$ and $k$ ($1 \leq n \leq 1000$, $0 \leq k \leq 1000$) — the number of skewers and the number of skewers from each side that are turned in one step.

第一行包含两个整数 $n$ 和 $k$（$1 \leq n \leq 1000$，$0 \leq k \leq 1000$）—— 分别表示烤肉签的总数，以及每一步中从每一侧翻转的烤肉签数量。

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.

第一行应包含整数 $l$ —— 米罗斯拉夫将全部 $n$ 个烤肉签翻转所需的最少操作次数。随后输出 $l$ 个介于 $1$ 到 $n$ 之间的整数，表示在对应步骤中需要翻转的烤肉签编号。

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.

在第一个例子中，第一次操作翻转烤肉签 $1$、$2$ 和 $3$，第二次操作翻转烤肉签 $4$、$5$、$6$ 和 $7$。

在第二个例子中，翻转烤肉签 $2$ 和 $5$ 也是正确的；但翻转烤肉签 $2$ 和 $4$，或 $1$ 和 $5$ 是错误的解法，因为经过这些操作后，烤肉签 $3$ 仍处于初始状态。