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.

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.

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.

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.

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