Ceil Divisions
Contest: Educational Codeforces Round 101 (Rated for Div. 2)
Difficulty rating: 1700
Tags: brute force, constructive algorithms, math, number theory

You have an array $$$a_1, a_2, \dots, a_n$$$ where $$$a_i = i$$$.

In one step, you can choose two indices $$$x$$$ and $$$y$$$ ($$$x \neq y$$$) and set $$$a_x = \left\lceil \frac{a_x}{a_y} \right\rceil$$$ (ceiling function).

Your goal is to make array $$$a$$$ consist of $$$n - 1$$$ ones and $$$1$$$ two in no more than $$$n + 5$$$ steps. Note that you don't have to minimize the number of steps.

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

The first and only line of each test case contains the single integer $$$n$$$ ($$$3 \le n \le 2 \cdot 10^5$$$) — the length of array $$$a$$$.

It's guaranteed that the sum of $$$n$$$ over test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output format:
For each test case, print the sequence of operations that will make $$$a$$$ as $$$n - 1$$$ ones and $$$1$$$ two in the following format: firstly, print one integer $$$m$$$ ($$$m \le n + 5$$$) — the number of operations; next print $$$m$$$ pairs of integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n$$$; $$$x \neq y$$$) ($$$x$$$ may be greater or less than $$$y$$$) — the indices of the corresponding operation.

It can be proven that for the given constraints it's always possible to find a correct sequence of operations.

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

Note:
In the first test case, you have array $$$a = [1, 2, 3]$$$. For example, you can do the following:

1. choose $$$3$$$, $$$2$$$: $$$a_3 = \left\lceil \frac{a_3}{a_2} \right\rceil = 2$$$ and array $$$a = [1, 2, 2]$$$;
2. choose $$$3$$$, $$$2$$$: $$$a_3 = \left\lceil \frac{2}{2} \right\rceil = 1$$$ and array $$$a = [1, 2, 1]$$$.

In the second test case, $$$a = [1, 2, 3, 4]$$$. For example, you can do the following:

1. choose $$$3$$$, $$$4$$$: $$$a_3 = \left\lceil \frac{3}{4} \right\rceil = 1$$$ and array $$$a = [1, 2, 1, 4]$$$;
2. choose $$$4$$$, $$$2$$$: $$$a_4 = \left\lceil \frac{4}{2} \right\rceil = 2$$$ and array $$$a = [1, 2, 1, 2]$$$;
3. choose $$$4$$$, $$$2$$$: $$$a_4 = \left\lceil \frac{2}{2} \right\rceil = 1$$$ and array $$$a = [1, 2, 1, 1]$$$.

Example input:
2
3
4

Example output:
2
3 2
3 2
3
3 4
4 2
4 2

Editorial:
There are many different approaches. We will describe a pretty optimal one.
Let's solve the problem recursively. Let's say we need to process segment $$$[1, x]$$$. If $$$x = 2$$$, we don't need to do anything. Otherwise, $$$x > 2$$$. Let's find the minimum $$$y$$$ such that $$$y \ge \left\lceil \frac{x}{y} \right\rceil$$$.
The chosen $$$y$$$ is convenient because it allows making $$$x$$$ equal to $$$1$$$ in two divisions (and it's the minimum number of divisions to get rid of $$$x$$$). Now we can, firstly, make all $$$z \in [y + 1, x)$$$ equal to $$$1$$$ in one step (by division on $$$x$$$) and then make $$$x$$$ equal to $$$1$$$ with two divisions on $$$y$$$.
As a result, we've spent $$$x - y + 1$$$ operations and can solve our task recursively for $$$[1, y]$$$. In total, we will spend $$$n - 2 + \mathit{\text{number_of_segments}}$$$ and since the segments are like $$$(\sqrt{x}, x]$$$, $$$(\sqrt{\sqrt{x}}, \sqrt{x}]$$$,... There will be at most $$$\log{(\log{n})}$$$ segments and $$$n + 3$$$ operations are enough for $$$n \le 2^{32}$$$.