Madhouse (Hard version)
Contest: Codeforces Round 612 (Div. 1)
Difficulty rating: 2800
Tags: brute force, constructive algorithms, hashing, interactive, math

This problem is different with easy version only by constraints on total answers length

It is an interactive problem

Venya joined a tour to the madhouse, in which orderlies play with patients the following game. Orderlies pick a string $$$s$$$ of length $$$n$$$, consisting only of lowercase English letters. The player can ask two types of queries:

- ? l r – ask to list all substrings of $$$s[l..r]$$$. Substrings will be returned in random order, and in every substring, all characters will be randomly shuffled.
- ! s – guess the string picked by the orderlies. This query can be asked exactly once, after that the game will finish. If the string is guessed correctly, the player wins, otherwise he loses.

The player can ask no more than $$$3$$$ queries of the first type.

To make it easier for the orderlies, there is an additional limitation: the total number of returned substrings in all queries of the first type must not exceed $$$\left\lceil 0.777(n+1)^2 \right\rceil$$$ ($$$\lceil x \rceil$$$ is $$$x$$$ rounded up).

Venya asked you to write a program, which will guess the string by interacting with the orderlies' program and acting by the game's rules.

Your program should immediately terminate after guessing the string using a query of the second type. In case your program guessed the string incorrectly, or it violated the game rules, it will receive verdict Wrong answer.

Note that in every test case the string is fixed beforehand and will not change during the game, which means that the interactor is not adaptive.

Input format:
First line contains number $$$n$$$ ($$$1 \le n \le 100$$$) — the length of the picked string.

Interaction format:
You start the interaction by reading the number $$$n$$$.

To ask a query about a substring from $$$l$$$ to $$$r$$$ inclusively ($$$1 \le l \le r \le n$$$), you should output

? l r

on a separate line. After this, all substrings of $$$s[l..r]$$$ will be returned in random order, each substring exactly once. In every returned substring all characters will be randomly shuffled.

In the case, if you ask an incorrect query, ask more than $$$3$$$ queries of the first type or there will be more than $$$\left\lceil 0.777(n+1)^2 \right\rceil$$$ substrings returned in total, you will receive verdict Wrong answer.

To guess the string $$$s$$$, you should output

! s

on a separate line.

After printing each query, do not forget to flush the output. Otherwise, you will get Idleness limit exceeded. To flush the output, you can 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.

If you received - (dash) as an answer to any query, you need to terminate your program with exit code 0 (for example, by calling exit(0)). This means that there was an error in the interaction protocol. If you don't terminate with exit code 0, you can receive any unsuccessful verdict.

Hack format

To hack a solution, use the following format:

The first line should contain one integer $$$n$$$ ($$$1 \le n \le 100$$$) — the length of the string, and the following line should contain the string $$$s$$$.

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

Example input:
4

a
aa
a

cb
b
c

c

Example output:
? 1 2

? 3 4

? 4 4

! aabc

Editorial:
Let's consider the solution that uses $$$2$$$ queries with the lengths $$$n$$$ and $$$n-1$$$ (it asks about too many substrings, so it will not pass all the tests, but it will help us further). Let's ask about substrings $$$[1..n]$$$ and $$$[1..n-1]$$$. For convenience, rearrange the letters in all strings in alphabetical order. Then note that all the suffixes of $$$S$$$ correspond to those strings that we get in the first case, and do not get in the second. Having found all such strings, we can easily find all suffixes of $$$S$$$ by looking at them from smaller ones to bigger.
For a complete solution we should first find the first $$$n / 2$$$ characters of the string by the solution described above. Then ask about the whole string. Let $$$cnt_{i, x}$$$ be the number of times that the symbol $$$x$$$ occurs in total in all substrings of length $$$i$$$ in the last query. Note that the symbol at position $$$j$$$ is counted in $$$cnt_{i, x}$$$ exactly $$$min(i, j, n - j + 1)$$$ times. Then for all $$$i$$$ from $$$1$$$ to $$$n / 2$$$, the value $$$cnt_{i + 1, x}$$$ - $$$cnt_{i,x}$$$ is equal to the number of times that $$$x$$$ occurs on positions with indices from $$$i + 1$$$ to $$$n - i$$$. Knowing these quantities, it is possible to find out how many times the element x occurs on positions $$$i$$$ and $$$n - i + 1$$$ in sum for each $$$i$$$ from $$$1$$$ to $$$n / 2$$$. Since we already know the first half of the string, it is not difficult to restore the character at the position $$$n - i + 1$$$, and therefore the entire string.
In total, we asked about the substrings $$$[1 .. n / 2]$$$, $$$[1..n / 2 - 1]$$$ and $$$[1..n]$$$, the total number of substrings received is $$$\approx \frac{(\frac{n + 1}{2})^2}{2}+ \frac{(\frac{n + 1}{2})^2}{2} + \frac{(n + 1)^2}{2} = 0.75 (n + 1)^2$$$, and that quite satisfies the limitations of the problem.