Macaque has taken you to his habitat, shown you his job, and even forced you to do his homework for him. While it might have not been easy for you, Macaque has grown to like you and even, possibly, harbour some gratitude towards you. However, all is not finished, for Macaque has to solve his most difficult problem yet. Solving this is a matter of life or death, and determines whether Macaque can finally reach enlightenment and retreat to solitude in an organ loft for the rest of eternity. Good luck. You are given two integers (n) and (m), and an array (b) of length (n) ((0 \le b_i \lt m)). It is guaranteed that every integer from (0) to (m-1) appears in (b) at least once. A group of (n) people are standing in a line, in fixed positions numbered (1, 2, \dots, n). They all want to sit down, but some are too socially awkward to do so immediately. For some arbitrary array (a) of length (n) ((0 \le a_i \lt n)), each person (i) has a social awkwardness level (a_i). Consider the seating process which evolves in discrete units of time (0, 1, 2, \dots) as follows: If (a_i = 0), person (i) will sit down at time (0). Otherwise ((a_i \gt 0)), person (i) will sit down at the beginning of a time unit (t) if both of the following conditions are satisfied: At least (a_i) people have already sat down strictly before time (t). At least one of their neighbors (person (i - 1) or (i + 1), if they exist) has already sat down strictly before time (t). At least (a_i) people have already sat down strictly before time (t). At least one of their neighbors (person (i - 1) or (i + 1), if they exist) has already sat down strictly before time (t). At the beginning of every unit of time, each standing person checks these conditions. All people for whom the conditions are satisfied sit down simultaneously , and then the process advances to the next unit of time.