Getting Zero

题意翻译

给定 $v$，可以将 $v$ 如下操作： 变成 $(v + 1) \% 32768$ 或者 $2\times v\%32768$， 求最少经过几次操作能将 $v$ 变回 $0$。

题目描述

Suppose you have an integer $ v $ . In one operation, you can: - either set $ v = (v + 1) \bmod 32768 $ - or set $ v = (2 \cdot v) \bmod 32768 $ . You are given $ n $ integers $ a_1, a_2, \dots, a_n $ . What is the minimum number of operations you need to make each $ a_i $ equal to $ 0 $ ?

输入输出格式

输入格式

The first line contains the single integer $ n $ ( $ 1 \le n \le 32768 $ ) — the number of integers. The second line contains $ n $ integers $ a_1, a_2, \dots, a_n $ ( $ 0 \le a_i < 32768 $ ).

输出格式

Print $ n $ integers. The $ i $ -th integer should be equal to the minimum number of operations required to make $ a_i $ equal to $ 0 $ .

输入输出样例

输入样例 #1

4
19 32764 10240 49

输出样例 #1

14 4 4 15

说明

Let's consider each $ a_i $ : - $ a_1 = 19 $ . You can, firstly, increase it by one to get $ 20 $ and then multiply it by two $ 13 $ times. You'll get $ 0 $ in $ 1 + 13 = 14 $ steps. - $ a_2 = 32764 $ . You can increase it by one $ 4 $ times: $ 32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0 $ . - $ a_3 = 10240 $ . You can multiply it by two $ 4 $ times: $ 10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0 $ . - $ a_4 = 49 $ . You can multiply it by two $ 15 $ times.

题意翻译

给定 $v$，可以将 $v$ 如下操作： 变成 $(v + 1) \% 32768$ 或者 $2\times v\%32768$， 求最少经过几次操作能将 $v$ 变回 $0$。