Guessing the Greatest (hard version)

The only difference between the easy and the hard version is the limit to the number of queries.

This is an interactive problem.

There is an array $$$a$$$ of $$$n$$$ different numbers. In one query you can ask the position of the second maximum element in a subsegment $$$a[l..r]$$$. Find the position of the maximum element in the array in no more than 20 queries.

A subsegment $$$a[l..r]$$$ is all the elements $$$a_l, a_{l + 1}, ..., a_r$$$. After asking this subsegment you will be given the position of the second maximum from this subsegment in the whole array.

You can ask queries by printing "? $$$l$$$ $$$r$$$" $$$(1 \leq l < r \leq n)$$$. The answer is the index of the second maximum of all elements $$$a_l, a_{l + 1}, ..., a_r$$$. Array $$$a$$$ is fixed beforehand and can't be changed in time of interaction.

You can output the answer by printing "! $$$p$$$", where $$$p$$$ is the index of the maximum element in the array.

You can ask no more than 20 queries. Printing the answer doesn't count as a query.

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

Hacks

To make a hack, use the following test format.

In the first line output a single integer $$$n$$$ $$$(2 \leq n \leq 10^5)$$$. In the second line output a permutation of $$$n$$$ integers $$$1$$$ to $$$n$$$. The position of $$$n$$$ in the permutation is the position of the maximum