A chef has collected data on the satisfaction level of his n dishes. Chef can cook any dish in 1 unit of time.
Like-time coefficient of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. time[i] * satisfaction[i].
Return the maximum sum of like-time coefficient that the chef can obtain after preparing some amount of dishes.
\n\nDishes can be prepared in any order and the chef can discard some dishes to get this maximum value.
\n\n\n
Example 1:
\n\n\nInput: satisfaction = [-1,-8,0,5,-9]\nOutput: 14\nExplanation: After Removing the second and last dish, the maximum total like-time coefficient will be equal to (-1*1 + 0*2 + 5*3 = 14).\nEach dish is prepared in one unit of time.\n\n
Example 2:
\n\n\nInput: satisfaction = [4,3,2]\nOutput: 20\nExplanation: Dishes can be prepared in any order, (2*1 + 3*2 + 4*3 = 20)\n\n\n
Example 3:
\n\n\nInput: satisfaction = [-1,-4,-5]\nOutput: 0\nExplanation: People do not like the dishes. No dish is prepared.\n\n\n
\n
Constraints:
\n\nn == satisfaction.length1 <= n <= 500-1000 <= satisfaction[i] <= 1000一个厨师收集了他 n 道菜的满意程度 satisfaction ，这个厨师做出每道菜的时间都是 1 单位时间。
一道菜的 「 like-time 系数 」定义为烹饪这道菜结束的时间（包含之前每道菜所花费的时间）乘以这道菜的满意程度，也就是 time[i]*satisfaction[i] 。
返回厨师在准备了一定数量的菜肴后可以获得的最大 like-time 系数 总和。
\n\n你可以按 任意 顺序安排做菜的顺序，你也可以选择放弃做某些菜来获得更大的总和。
\n\n\n\n
示例 1：
\n\n\n输入：satisfaction = [-1,-8,0,5,-9]\n输出：14\n解释：去掉第二道和最后一道菜，最大的 like-time 系数和为 (-1*1 + 0*2 + 5*3 = 14) 。每道菜都需要花费 1 单位时间完成。\n\n
示例 2：
\n\n\n输入：satisfaction = [4,3,2]\n输出：20\n解释：可以按照任意顺序做菜 (2*1 + 3*2 + 4*3 = 20)\n\n\n
示例 3：
\n\n\n输入：satisfaction = [-1,-4,-5]\n输出：0\n解释：大家都不喜欢这些菜，所以不做任何菜就可以获得最大的 like-time 系数。\n\n\n
\n\n
提示：
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