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

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.

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.