The hat
Contest: Codeforces Round 503 (by SIS, Div. 1)
Difficulty rating: 2000
Tags: binary search, interactive

This is an interactive problem.

Imur Ishakov decided to organize a club for people who love to play the famous game «The hat». The club was visited by n students, where n is even. Imur arranged them all in a circle and held a draw to break the students in pairs, but something went wrong. The participants are numbered so that participant i and participant i + 1 (1 ≤ i ≤ n - 1) are adjacent, as well as participant n and participant 1. Each student was given a piece of paper with a number in such a way, that for every two adjacent students, these numbers differ exactly by one. The plan was to form students with the same numbers in a pair, but it turned out that not all numbers appeared exactly twice.

As you know, the most convenient is to explain the words to the partner when he is sitting exactly across you. Students with numbers i and $$i + \frac{n}{2}$$ sit across each other. Imur is wondering if there are two people sitting across each other with the same numbers given. Help him to find such pair of people if it exists.

You can ask questions of form «which number was received by student i?», and the goal is to determine whether the desired pair exists in no more than 60 questions.

Input format:
At the beginning the even integer n (2 ≤ n ≤ 100 000) is given — the total number of students.

You are allowed to ask no more than 60 questions.

Output format:
To ask the question about the student i (1 ≤ i ≤ n), you should print «? i». Then from standard output you can read the number ai received by student i ( - 109 ≤ ai ≤ 109).

When you find the desired pair, you should print «! i», where i is any student who belongs to the pair (1 ≤ i ≤ n). If you determined that such pair doesn't exist, you should output «! -1». In both cases you should immediately terminate the program.

The query that contains your answer is not counted towards the limit of 60 queries.

Please make sure to flush the standard output after each command. For example, in C++ use function fflush(stdout), in Java call System.out.flush(), in Pascal use flush(output) and stdout.flush() for Python language.

Hacking

Use the following format for hacking:

In the first line, print one even integer n (2 ≤ n ≤ 100 000) — the total number of students.

In the second line print n integers ai ( - 109 ≤ ai ≤ 109) separated by spaces, where ai is the number to give to i-th student. Any two adjacent elements, including n and 1, must differ by 1 or  - 1.

The hacked solution will not have direct access to the sequence ai.

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

Note:
Input-output in statements illustrates example interaction.

In the first sample the selected sequence is 1, 2, 1, 2, 3, 4, 3, 2

In the second sample the selection sequence is 1, 2, 3, 2, 1, 0.

Example input:
8
2
2

Example output:
? 4
? 8
! 4

Example input:
6
1
2
3
2
1
0

Example output:
? 1
? 2
? 3
? 4
? 5
? 6
! -1

Editorial:
Let a(i) be a number given to the i-th student. Let's introduce function $$b(i) = a(i) - a(i + \frac{n}{2})$$. As n is even, $$\frac{\pi}{2}$$ is integer, and b(i) is defined correctly. Notice two facts: first, $$b(i) = - b(i + \frac{n}{2})$$, and second, $$| b ( i ) - b ( i + 1 ) | \in \{ - 2, 0, 2 \}$$. The problem is to find an i such that b(i) = 0.
Second note leads us to observation that all b(i) have the same oddity. So, let's find b(0), and if it is odd, then there is no answer, and we need to print  - 1 — all b(i) have the same oddity (odd), and zero, as a number of different oddity, won't appear in b(i).
Otherwise, suppose we know b(0), and it is equal to x (without loss of generality x > 0). Notice that $$b \left( 0 + \frac { n } { 2 } \right) = - b ( 0 ) = - x$$. As we remember from second observation that all b(i) have the same oddity, and neighboring numbers differ by no more than 2, we can use discrete continuity.
Lemma: if you have two indices i and j with values b(i) and b(j), then on segment between i and j there are all values with the same oddity from min(b(i), b(j)) to max(b(i), b(j)). Indeed, as neighboring b differ by no more than 2, we couldn't skip any number with the same oddity.
Now we can use a binary search. At the start l = 0, $$r = \frac{n}{2}$$, and values b(l) and b(r) are even numbers with the different signs. By lemma, on segment between l and r there is a zero that we want to find. Let's take $$m = \frac{l + r}{2}$$. If b(m) = 0, we found the answer. Otherwise, based on sign of b(m) we can replace one of l, r in binary search to m, and reduce the problem to the same with smaller size. There will be no more than $$\log n$$ iterations, and as calculation of b(i) requires two queries, we solved the problem in $$2\log n$$ queries.