CF1583D.Omkar and the Meaning of Life

It turns out that the meaning of life is a permutation $p_1, p_2, \ldots, p_n$ of the integers $1, 2, \ldots, n$ ($2 \leq n \leq 100$). Omkar, having created all life, knows this permutation, and will allow you to figure it out using some queries.

A query consists of an array $a_1, a_2, \ldots, a_n$ of integers between $1$ and $n$. $a$ is not required to be a permutation. Omkar will first compute the pairwise sum of $a$ and $p$, meaning that he will compute an array $s$ where $s_j = p_j + a_j$ for all $j = 1, 2, \ldots, n$. Then, he will find the smallest index $k$ such that $s_k$ occurs more than once in $s$, and answer with $k$. If there is no such index $k$, then he will answer with $0$.

You can perform at most $2n$ queries. Figure out the meaning of life $p$.

Interaction

Start the interaction by reading single integer $n$ ($2 \leq n \leq 100$) — the length of the permutation $p$.

You can then make queries. A query consists of a single line "$? \enspace a_1 \enspace a_2 \enspace \ldots \enspace a_n$" ($1 \leq a_j \leq n$).

The answer to each query will be a single integer $k$ as described above ($0 \leq k \leq n$).

After making a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

• fflush(stdout) or cout.flush() in C++;
• System.out.flush() in Java;
• flush(output) in Pascal;
• stdout.flush() in Python;
• see documentation for other languages.

To output your answer, print a single line "$! \enspace p_1 \enspace p_2 \enspace \ldots \enspace p_n$" then terminate.

You can make at most $2n$ queries. Outputting the answer does not count as a query.

Hack Format

To hack, first output a line containing $n$ ($2 \leq n \leq 100$), then output another line containing the hidden permutation $p_1, p_2, \ldots, p_n$ of numbers from $1$ to $n$.

事实证明，生命的意义是一个 $1, 2, \ldots, n$ 的排列 $p_1, p_2, \ldots, p_n$（其中 $2 \leq n \leq 100$）。奥姆卡尔（Omkar）创造了全部生命，因此他知道这个排列，并允许你通过若干次查询来推断出它。

一次查询由一个长度为 $n$ 的数组 $a_1, a_2, \ldots, a_n$ 构成，其中每个 $a_j$ 是介于 $1$ 和 $n$ 之间的整数（$a$ 不必是排列）。奥姆卡尔首先计算 $a$ 与 $p$ 的逐项和，即构造一个数组 $s$，使得对所有 $j = 1, 2, \ldots, n$，均有 $s_j = p_j + a_j$。然后，他找出最小的下标 $k$，使得 $s_k$ 在数组 $s$ 中出现次数超过一次，并将该 $k$ 作为回答返回；若不存在这样的 $k$，则回答为 $0$。

你最多可进行 $2n$ 次查询。请推断出生命的意义——即排列 $p$。

交互流程

首先读入一个整数 $n$（$2 \leq n \leq 100$），表示排列 $p$ 的长度。

随后你可以发起查询。每次查询格式为一行：? a_1 a_2 … a_n（其中 $1 \leq a_j \leq n$）。

每次查询的响应为一个整数 $k$（$0 \leq k \leq n$），其含义如上所述。

每次查询后，请务必输出换行符并刷新输出缓冲区，否则将收到 “Idleness limit exceeded” 错误。为此，请使用以下方式：

• C++ 中使用 fflush(stdout) 或 cout.flush()；
• Java 中使用 System.out.flush()；
• Pascal 中使用 flush(output)；
• Python 中使用 stdout.flush()；
• 其他语言请查阅相应文档。

当你得出答案后，请输出一行：! p_1 p_2 … p_n，然后终止程序。

你最多可进行 $2n$ 次查询。输出最终答案不计入查询次数。

Hack 格式

用于 Hack 时，首先输出一行包含整数 $n$（$2 \leq n \leq 100$），然后输出另一行，包含隐藏的排列 $p_1, p_2, \ldots, p_n$（由 $1$ 到 $n$ 的整数组成）。

In the sample, the hidden permutation $p$ is $[3, 2, 1, 5, 4]$. Three queries were made.

The first query is $a = [4, 4, 2, 3, 2]$. This yields $s = [3 + 4, 2 + 4, 1 + 2, 5 + 3, 4 + 2] = [7, 6, 3, 8, 6]$. $6$ is the only number that appears more than once, and it appears first at index $2$, making the answer to the query $2$.

The second query is $a = [3, 5, 1, 5, 5]$. This yields $s = [3 + 3, 2 + 5, 1 + 1, 5 + 5, 4 + 5] = [6, 7, 2, 10, 9]$. There are no numbers that appear more than once here, so the answer to the query is $0$.

The third query is $a = [5, 2, 4, 3, 1]$. This yields $s = [3 + 5, 2 + 2, 1 + 4, 5 + 3, 4 + 1] = [8, 4, 5, 8, 5]$. $5$ and $8$ both occur more than once here. $5$ first appears at index $3$, while $8$ first appears at index $1$, and $1 \lt 3$, making the answer to the query $1$.

Note that the sample is only meant to provide an example of how the interaction works; it is not guaranteed that the above queries represent a correct strategy with which to determine the answer.

在样例中，隐藏的排列 $p$ 为 $[3, 2, 1, 5, 4]$。共进行了三次查询。

第一次查询为 $a = [4, 4, 2, 3, 2]$。由此得到 $s = [3 + 4, 2 + 4, 1 + 2, 5 + 3, 4 + 2] = [7, 6, 3, 8, 6]$。其中仅数字 $6$ 出现超过一次，且首次出现在索引 $2$ 处，因此该查询的答案为 $2$。

第二次查询为 $a = [3, 5, 1, 5, 5]$。由此得到 $s = [3 + 3, 2 + 5, 1 + 1, 5 + 5, 4 + 5] = [6, 7, 2, 10, 9]$。此处没有任何数字出现超过一次，因此该查询的答案为 $0$。

第三次查询为 $a = [5, 2, 4, 3, 1]$。由此得到 $s = [3 + 5, 2 + 2, 1 + 4, 5 + 3, 4 + 1] = [8, 4, 5, 8, 5]$。其中 $5$ 和 $8$ 均出现超过一次。$5$ 首次出现在索引 $3$ 处，而 $8$ 首次出现在索引 $1$ 处，且 $1 \lt 3$，因此该查询的答案为 $1$。

注意：该样例仅用于说明交互方式；不能保证上述查询构成一种能正确确定答案的有效策略。