An element (b_i) ((1\le i\le m)) in an array (b_1, b_2, \ldots, b_m) is a local minimum if at least one of the following holds: (2\le i\le m - 1) and (b_i < b_{i - 1}) and (b_i < b_{i + 1}), or (i = 1) and (b_1 < b_2), or (i = m) and (b_m < b_{m - 1}). Similarly, an element (b_i) ((1\le i\le m)) in an array (b_1, b_2, \ldots, b_m) is a local maximum if at least one of the following holds: (2\le i\le m - 1) and (b_i > b_{i - 1}) and (b_i > b_{i + 1}), or (i = 1) and (b_1 > b_2), or (i = m) and (b_m > b_{m - 1}). Note that local minima and maxima are not defined for arrays with only one element. There is a hidden permutation(^{\text{∗}}) (p) of length (n). The following two operations are applied to permutation (p) alternately, starting from operation 1, until there is only one element left in (p): Operation 1 — remove all elements of (p) which are not local minima. Operation 2 — remove all elements of (p) which are not local maxima. More specifically, operation 1 is applied during every odd iteration, and operation 2 is applied during every even iteration, until there is only one element left in (p). For each index (i) ((1\le i\le n)), let (a_i) be the iteration number that element (p_i) is removed, or (-1) if it was never removed. It can be proven that there will be only one element left in (p) after at most (\lceil \log_2 n\rceil) iterations (in other words, (a_i \le \lceil \log_2 n\rceil)). You are given the array (a_1, a_2, \ldots, a_n). Your task is to construct any permutation (p) of (n) elements that satisfies array (a). (^{\text{∗}})A permutation of length (n) is an array consisting of (n) distinct integers from (1) to (n) in arbitrary order. For example, (2,3,1,5,4) is a permutation, but (1,2,2) is not a permutation ((2) appears twice in the array), and $$