In Search of Truth (Easy Version)
Contest: Codeforces Round 878 (Div. 3)
Difficulty rating: 2200
Tags: constructive algorithms, interactive, math, meet-in-the-middle, probabilities

The only difference between easy and hard versions is the maximum number of queries. In this version, you are allowed to ask at most $$$2023$$$ queries.

This is an interactive problem.

You are playing a game. The circle is divided into $$$n$$$ sectors, sectors are numbered from $$$1$$$ to $$$n$$$ in some order. You are in the adjacent room and do not know either the number of sectors or their numbers. There is also an arrow that initially points to some sector. Initially, the host tells you the number of the sector to which the arrow points. After that, you can ask the host to move the arrow $$$k$$$ sectors counterclockwise or clockwise at most $$$2023$$$ times. And each time you are told the number of the sector to which the arrow points.

Your task is to determine the integer $$$n$$$ — the number of sectors in at most $$$2023$$$ queries.

It is guaranteed that $$$1 \le n \le 10^6$$$.

Input format:
The input consists of a single integer $$$x$$$ ($$$1 \le x \le n$$$) — the number of the initial sector.

Output format:
After you determine the integer $$$n$$$ — the number of sectors, you should output "! n" ($$$1 \le n \le 10^6$$$). After that the program should immediately terminate.

Note that, printing the answer does not count as a query.

It is guaranteed that the integer $$$n$$$ and the numbers of the sectors are fixed initially and will not be changed by the jury program depending on the queries.

Interaction format:
After reading the description of the input, you may ask queries. Queries can be of two types:

1. "+ k" ($$$0 \le k \le 10^9$$$) — ask to move the arrow $$$k$$$ sectors clockwise.
2. "- k" ($$$0 \le k \le 10^9$$$) — ask to move the arrow $$$k$$$ sectors counterclockwise.

After each query, you should read an integer $$$x$$$ ($$$1 \le x \le n$$$) — the number of the current sector to which the arrow points.

You are allowed to make at most $$$2023$$$ queries in total.

If you make too many queries, you will get Wrong answer.

After printing a query or the answer, do not forget to output a the 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 the documentation for other languages.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
Hacks

To hack, use the following test format.

In the first line, output a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$) — the number of sectors.

In the second line, output $$$n$$$ different integers $$$1 \le a_1, a_2, \dots, a_n \le n$$$ — the numbers of the sectors in clockwise order, the arrow initially points to the sector with the number $$$a_1$$$.

Example input:
1

5

6

7

2

10

9

8

4

3

1

Example output:
+ 1

+ 1

+ 1

+ 1

+ 1

+ 1

+ 1

+ 1

+ 1

+ 1

! 10

Editorial:
Let $$$a_1, a_2, \dots, a_n$$$ be the numbers of the sectors in clockwise order, and let the arrow initially point to the sector with number $$$a_1$$$.
First, let's make $$$999$$$ queries of "+ 1", then we will know the numbers of $$$1000$$$ consecutive sectors. If $$$n < 1000$$$, then the number of the first query that gives the answer $$$a_1$$$ is the desired $$$n$$$.
If we did not find $$$n$$$, this means that $$$n \ge 1000$$$. Let's save $$$a_1, a_2, \dots, a_{1000}$$$. Now we will make queries of "+ 1000" until we get one of the numbers $$$a_1, a_2, \dots, a_{1000}$$$ as the answer. Note that we will need no more than $$$1000$$$ queries of this type, after which it is easy to determine the number $$$n$$$.
Thus, we can determine the number $$$n$$$ in no more than $$$999 + 1000 = 1999 \le 2023$$$ queries.