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.

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.