Everyone is a Winner!
Contest: Codeforces Round 603 (Div. 2)
Difficulty rating: 1400
Tags: binary search, math, meet-in-the-middle, number theory

On the well-known testing system MathForces, a draw of $$$n$$$ rating units is arranged. The rating will be distributed according to the following algorithm: if $$$k$$$ participants take part in this event, then the $$$n$$$ rating is evenly distributed between them and rounded to the nearest lower integer, At the end of the drawing, an unused rating may remain — it is not given to any of the participants.

For example, if $$$n = 5$$$ and $$$k = 3$$$, then each participant will recieve an $$$1$$$ rating unit, and also $$$2$$$ rating units will remain unused. If $$$n = 5$$$, and $$$k = 6$$$, then none of the participants will increase their rating.

Vasya participates in this rating draw but does not have information on the total number of participants in this event. Therefore, he wants to know what different values of the rating increment are possible to get as a result of this draw and asks you for help.

For example, if $$$n=5$$$, then the answer is equal to the sequence $$$0, 1, 2, 5$$$. Each of the sequence values (and only them) can be obtained as $$$\lfloor n/k \rfloor$$$ for some positive integer $$$k$$$ (where $$$\lfloor x \rfloor$$$ is the value of $$$x$$$ rounded down): $$$0 = \lfloor 5/7 \rfloor$$$, $$$1 = \lfloor 5/5 \rfloor$$$, $$$2 = \lfloor 5/2 \rfloor$$$, $$$5 = \lfloor 5/1 \rfloor$$$.

Write a program that, for a given $$$n$$$, finds a sequence of all possible rating increments.

Input format:
The first line contains integer number $$$t$$$ ($$$1 \le t \le 10$$$) — the number of test cases in the input. Then $$$t$$$ test cases follow.

Each line contains an integer $$$n$$$ ($$$1 \le n \le 10^9$$$) — the total number of the rating units being drawn.

Output format:
Output the answers for each of $$$t$$$ test cases. Each answer should be contained in two lines.

In the first line print a single integer $$$m$$$ — the number of different rating increment values that Vasya can get.

In the following line print $$$m$$$ integers in ascending order — the values of possible rating increments.

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

Example input:
4
5
11
1
3

Example output:
4
0 1 2 5 
6
0 1 2 3 5 11 
2
0 1 
3
0 1 3

Editorial:
There are two approaches to solving this problem.
Mathematical Solution
Note that the answer will always contain the numbers $$$0 \le x < \lfloor \sqrt{n} \rfloor$$$. You can verify this by solving the equation $$$\lfloor \frac{n}{k} \rfloor = x$$$, equivalent to the inequality $$$x \le \frac{n}{k} < x + 1$$$, for integer values ​​of $$$k$$$. The solution to this double inequality is the interval $$$k \in \left (\frac{n}{x + 1};\; \frac{n}{x} \right]$$$, whose length is $$$\frac{n}{x^2 + x}$$$. For $$$x < \lfloor \sqrt{n} \rfloor$$$ $$$\frac{n}{x^2 + x} > 1$$$, and on an interval of length greater than 1 there is always a whole solution $$$k = \lfloor \frac{n}{x} \rfloor$$$, so all integers $$$0 \le x < \lfloor \sqrt {n} \rfloor$$$ belong to the answer.
Note that we no longer need to iterate over the values ​​of $$$k > \lfloor \sqrt{n} \rfloor$$$, because these numbers always correspond to the values ​​$$$0 \le x < \lfloor \sqrt {n} \rfloor$$$. Thus, it is possible, as in a naive solution, to iterate over all the values ​​of $$$k$$$ upto $$$\lfloor \sqrt{n} \rfloor$$$ and add $$$x = \lfloor \frac{n}{k} \rfloor$$$ to the answer. It remains only to carefully handle the case $$$k = \lfloor \sqrt{n} \rfloor$$$.
Total complexity of the solution: $$$\mathcal{O} (\sqrt{n} \log n)$$$ or $$$\mathcal{O} (\sqrt{n})$$$
Algorithmic Solution
In the problem, it could be assumed that there are not so many numbers in the answer (after all, they still need to be printed, which takes the majority of the program execution time). Obviously, $$$n$$$ always belongs to the answer. Note that as $$$k$$$ increases, the value of $$$x = \lfloor \frac{n}{k} \rfloor$$$ decreases. Thus, using a binary search, you can find the smallest value of $$$k'$$$ such that $$$\frac{n}{k'} < x$$$. Value $$$x' = \frac{n}{k'}$$$ will be the previous one for $$$x$$$ in the answer.
Total complexity of the solution: $$$\mathcal{O} (\sqrt{n} \log n)$$$