CF1299E.So Mean

This problem is interactive.

We have hidden a permutation $p_1, p_2, \dots, p_n$ of numbers from $1$ to $n$ from you, where $n$ is even. You can try to guess it using the following queries:

$?$ $k$ $a_1$ $a_2$ $\dots$ $a_k$.

In response, you will learn if the average of elements with indexes $a_1, a_2, \dots, a_k$ is an integer. In other words, you will receive $1$ if $\frac{p_{a_1} + p_{a_2} + \dots + p_{a_k}}{k}$ is integer, and $0$ otherwise.

You have to guess the permutation. You can ask not more than $18n$ queries.

Note that permutations $[p_1, p_2, \dots, p_k]$ and $[n + 1 - p_1, n + 1 - p_2, \dots, n + 1 - p_k]$ are indistinguishable. Therefore, you are guaranteed that $p_1 \le \frac{n}{2}$.

Note that the permutation $p$ is fixed before the start of the interaction and doesn't depend on your queries. In other words, interactor is not adaptive.

Note that you don't have to minimize the number of queries.

本题为交互式问题。

我们隐藏了一个由 $1$ 到 $n$ 构成的排列 $p_1, p_2, \dots, p_n$，其中 $n$ 为偶数。你可以通过以下查询尝试猜出该排列：

$?$ $k$ $a_1$ $a_2$ $\dots$ $a_k$。

作为回应，你将得知下标为 $a_1, a_2, \dots, a_k$ 的元素的平均值是否为整数。换言之，若 $\frac{p_{a_1} + p_{a_2} + \dots + p_{a_k}}{k}$ 是整数，则返回 $1$；否则返回 $0$。

你需要猜出该排列。你至多可进行 $18n$ 次查询。

注意：排列 $[p_1, p_2, \dots, p_k]$ 与 $[n + 1 - p_1, n + 1 - p_2, \dots, n + 1 - p_k]$ 在查询下不可区分。因此，题目保证 $p_1 \le \frac{n}{2}$。

注意：排列 $p$ 在交互开始前即已固定，且不依赖于你的查询。换言之，交互器是非自适应的（non-adaptive）。

注意：你无需最小化查询次数。

The first line contains a single integer $n$ ($2 \le n \le 800$, $n$ is even).

第一行包含一个整数 $n$（$2 \le n \le 800$，且 $n$ 为偶数）。