{"trustable":true,"prependHtml":"\u003cstyle type\u003d\u0027text/css\u0027\u003e\n .input, .output {\n border: 1px solid #888888;\n }\n .output {\n margin-bottom: 1em;\n position: relative;\n top: -1px;\n }\n .output pre, .input pre {\n background-color: #EFEFEF;\n line-height: 1.25em;\n margin: 0;\n padding: 0.25em;\n }\n \u003c/style\u003e\n \u003clink rel\u003d\"stylesheet\" href\u003d\"//codeforces.org/s/96598/css/problem-statement.css\" type\u003d\"text/css\" /\u003e\n\u003cscript\u003e\n window.katexOptions \u003d {\n delimiters: [\n {left: \u0027$$$$$$\u0027, right: \u0027$$$$$$\u0027, display: true},\n {left: \u0027$$$\u0027, right: \u0027$$$\u0027, display: false},\n {left: \u0027$$\u0027, right: \u0027$$\u0027, display: true},\n {left: \u0027$\u0027, right: \u0027$\u0027, display: false}\n ]\n };\n\u003c/script\u003e\n","sections":[{"title":"","value":{"format":"HTML","content":"\u003cp\u003eLong story short, shashlik is Miroslav\u0027s favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.\u003c/p\u003e\u003cp\u003eThis time Miroslav laid out $$$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$$$ (if they exist). \u003c/p\u003e\u003cp\u003eFor example, let $$$n \u003d 6$$$ and $$$k \u003d 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$$$ will be in the initial position (because it is turned again).\u003c/p\u003e\u003cp\u003eAs we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all $$$n$$$ skewers with the minimal possible number of actions. For example, for the above example $$$n \u003d 6$$$ and $$$k \u003d 1$$$, two turnings are sufficient: he can turn over skewers number $$$2$$$ and $$$5$$$.\u003c/p\u003e\u003cp\u003eHelp Miroslav turn over all $$$n$$$ skewers.\u003c/p\u003e"}},{"title":"Input","value":{"format":"HTML","content":"\u003cp\u003eThe first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \\leq n \\leq 1000$$$, $$$0 \\leq k \\leq 1000$$$)\u0026nbsp;— the number of skewers and the number of skewers from each side that are turned in one step.\u003c/p\u003e"}},{"title":"Output","value":{"format":"HTML","content":"\u003cp\u003eThe first line should contain integer $$$l$$$\u0026nbsp;— the minimum number of actions needed by Miroslav to turn over all $$$n$$$ skewers. After than print $$$l$$$ integers from $$$1$$$ to $$$n$$$ denoting the number of the skewer that is to be turned over at the corresponding step.\u003c/p\u003e"}},{"title":"Examples","value":{"format":"HTML","content":"\u003ctable class\u003d\u0027vjudge_sample\u0027\u003e\n\u003cthead\u003e\n \u003ctr\u003e\n \u003cth\u003eInput\u003c/th\u003e\n \u003cth\u003eOutput\u003c/th\u003e\n \u003c/tr\u003e\n\u003c/thead\u003e\n\u003ctbody\u003e\n \u003ctr\u003e\n \u003ctd\u003e\u003cpre\u003e7 2\n\u003c/pre\u003e\u003c/td\u003e\n \u003ctd\u003e\u003cpre\u003e2\n1 6 \n\u003c/pre\u003e\u003c/td\u003e\n \u003c/tr\u003e\n\u003c/tbody\u003e\n\u003c/table\u003e\n"}},{"title":"","value":{"format":"HTML","content":"\u003ctable class\u003d\u0027vjudge_sample\u0027\u003e\n\u003cthead\u003e\n \u003ctr\u003e\n \u003cth\u003eInput\u003c/th\u003e\n \u003cth\u003eOutput\u003c/th\u003e\n \u003c/tr\u003e\n\u003c/thead\u003e\n\u003ctbody\u003e\n \u003ctr\u003e\n \u003ctd\u003e\u003cpre\u003e5 1\n\u003c/pre\u003e\u003c/td\u003e\n \u003ctd\u003e\u003cpre\u003e2\n1 4 \n\u003c/pre\u003e\u003c/td\u003e\n \u003c/tr\u003e\n\u003c/tbody\u003e\n\u003c/table\u003e\n"}},{"title":"Note","value":{"format":"HTML","content":"\u003cp\u003eIn the first example the first operation turns over skewers $$$1$$$, $$$2$$$ and $$$3$$$, the second operation turns over skewers $$$4$$$, $$$5$$$, $$$6$$$ and $$$7$$$.\u003c/p\u003e\u003cp\u003eIn the second example it is also correct to turn over skewers $$$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.\u003c/p\u003e"}}]}