Strange Device
Contest: Good Bye 2019
Difficulty rating: 1900
Tags: constructive algorithms, interactive, math, sortings

This problem is interactive.

We have hidden an array $$$a$$$ of $$$n$$$ pairwise different numbers (this means that no two numbers are equal). You can get some information about this array using a new device you just ordered on Amazon.

This device can answer queries of the following form: in response to the positions of $$$k$$$ different elements of the array, it will return the position and value of the $$$m$$$-th among them in the ascending order.

Unfortunately, the instruction for the device was lost during delivery. However, you remember $$$k$$$, but don't remember $$$m$$$. Your task is to find $$$m$$$ using queries to this device.

You can ask not more than $$$n$$$ queries.

Note that the array $$$a$$$ and number $$$m$$$ are fixed before the start of the interaction and don't depend on your queries. In other words, interactor is not adaptive.

Note that you don't have to minimize the number of queries, and you don't need to guess array $$$a$$$. You just have to guess $$$m$$$.

Input format:
The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1\le k < n \le 500$$$) — the length of the array and the number of the elements in the query.

It is guaranteed that number $$$m$$$ satisfies $$$1\le m \le k$$$, elements $$$a_1, a_2, \dots, a_n$$$ of the array satisfy $$$0\le a_i \le 10^9$$$, and all of them are different.

Interaction format:
You begin the interaction by reading $$$n$$$ and $$$k$$$.

To ask a question about elements on positions $$$x_1, x_2, \dots, x_k$$$, in a separate line output

$$$?$$$ $$$x_1$$$ $$$x_2$$$ $$$x_3$$$ ... $$$x_k$$$

Numbers in the query have to satisfy $$$1 \le x_i \le n$$$, and all $$$x_i$$$ have to be different. Don't forget to 'flush', to get the answer.

In response, you will receive two integers $$$pos$$$ and $$$a_{pos}$$$ — the position in the array $$$a$$$ of the $$$m$$$-th in ascending order element among $$$a_{x_1}, a_{x_2}, \dots, a_{x_k}$$$, and the element on this position.

In case your query is invalid or you asked more than $$$n$$$ queries, the program will print $$$-1$$$ and will finish interaction. You will receive a Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

When you determine $$$m$$$, output

$$$!$$$ $$$m$$$

After printing 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.

Hack format

For the hacks use the following format:

The first line has to contain three integers $$$n, k, m$$$ ($$$1 \le m \le k < n \le 500$$$) — the length of the array, number of elements in the query, and which in the ascending order number the device returns.

In the next line output $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0\le a_i \le 10^9$$$) — the elements of the array. They have to be pairwise different.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the example, $$$n = 4$$$, $$$k = 3$$$, $$$m = 3$$$, $$$a = [2, 0, 1, 9]$$$.

Example input:
4 3
4 9
4 9
4 9
1 2

Example output:
? 2 3 4
? 1 3 4
? 1 2 4
? 1 2 3
! 3

Editorial:
We will show how to guess $$$m$$$ with $$$k+1\le n$$$ queries.
Let's leave only first $$$k+1$$$ elements of the array. We will ask $$$k+1$$$ queries: $$$i$$$-th question — about all elements from $$$1$$$-th to $$$k+1$$$-th, except $$$i$$$-th. Denote elements from $$$1$$$-th to $$$k+1$$$-th in decreasing order as $$$b_1 < b_2 < \dots < b_{k+1}$$$. Then, if we throw out element which is larger than $$$b_m$$$, then $$$b_m$$$ is the $$$m$$$-th largest among remaining elements. If we throw out element which is smaller of equal to $$$b_m$$$, then $$$b_{m+1}$$$ is the $$$m$$$-th largest among remaining elements.
Therefore, among answers to our $$$k+1$$$ queries we will have $$$m$$$ times element $$$b_{m+1}$$$ and $$$k+1-m$$$ times element $$$b_m$$$. As $$$b_{m+1}>b_m$$$, we can first find $$$b_{m+1}$$$ (as the largest of all answers to the queries), and after that we can find $$$m$$$ as the number of times $$$b_{m+1}$$$ meets among these answers!