Neko Performs Cat Furrier Transform
Contest: Codeforces Round 554 (Div. 2)
Difficulty rating: 1300
Tags: bitmasks, constructive algorithms, dfs and similar, math

Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.

Assume that we have a cat with a number $$$x$$$. A perfect longcat is a cat with a number equal $$$2^m - 1$$$ for some non-negative integer $$$m$$$. For example, the numbers $$$0$$$, $$$1$$$, $$$3$$$, $$$7$$$, $$$15$$$ and so on are suitable for the perfect longcats.

In the Cat Furrier Transform, the following operations can be performed on $$$x$$$:

- (Operation A): you select any non-negative integer $$$n$$$ and replace $$$x$$$ with $$$x \oplus (2^n - 1)$$$, with $$$\oplus$$$ being a bitwise XOR operator.
- (Operation B): replace $$$x$$$ with $$$x + 1$$$.

The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.

Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most $$$40$$$ operations. Can you help Neko writing a transformation plan?

Note that it is not required to minimize the number of operations. You just need to use no more than $$$40$$$ operations.

Input format:
The only line contains a single integer $$$x$$$ ($$$1 \le x \le 10^6$$$).

Output format:
The first line should contain a single integer $$$t$$$ ($$$0 \le t \le 40$$$) — the number of operations to apply.

Then for each odd-numbered operation print the corresponding number $$$n_i$$$ in it. That is, print $$$\lceil \frac{t}{2} \rceil$$$ integers $$$n_i$$$ ($$$0 \le n_i \le 30$$$), denoting the replacement $$$x$$$ with $$$x \oplus (2^{n_i} - 1)$$$ in the corresponding step.

If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.

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

Note:
In the first test, one of the transforms might be as follows: $$$39 \to 56 \to 57 \to 62 \to 63$$$. Or more precisely:

1. Pick $$$n = 5$$$. $$$x$$$ is transformed into $$$39 \oplus 31$$$, or $$$56$$$.
2. Increase $$$x$$$ by $$$1$$$, changing its value to $$$57$$$.
3. Pick $$$n = 3$$$. $$$x$$$ is transformed into $$$57 \oplus 7$$$, or $$$62$$$.
4. Increase $$$x$$$ by $$$1$$$, changing its value to $$$63 = 2^6 - 1$$$.

In the second and third test, the number already satisfies the goal requirement.

Example input:
39

Example output:
4
5 3

Example input:
1

Example output:
0

Editorial:
There are various greedy solutions possible. I'll only cover one of them.
We'll perform a loop as follows:
• If $$$x \le 1$$$, stop the process (since it's already correct).
• If $$$x = 2^m$$$ ($$$m \ge 1$$$), perform operation A with $$$2^m-1$$$. Obviously, the resulting $$$x$$$ will be $$$2^{m+1}-1$$$, which satisfies the criteria, so we stop the process.
• Otherwise, let's denote $$$MSB(x)$$$ as the position/exponential of the most significant bit of $$$x$$$ (for example, $$$MSB(4) = MSB(7) = 2$$$). We'll perform operation A with $$$2^{MSB(x)+1} - 1$$$. If $$$x$$$ after this phase is not a valid number, we'll then perform operation B and return to the beginning of the loop.
This will never take more than $$$40$$$ queries, since each iteration removes the most significant bit (and it would never reappear after later steps), and $$$x$$$ can only have at most $$$20$$$ bits initially.
Also, it is possible to solve this problem by finding exactly shortest operation sequence by using a BFS. However, this was not required.