Prefix Flip (Easy Version)
Contest: Codeforces Round 658 (Div. 1)
Difficulty rating: 1300
Tags: constructive algorithms, data structures, strings

This is the easy version of the problem. The difference between the versions is the constraint on $$$n$$$ and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings $$$a$$$ and $$$b$$$ of length $$$n$$$ (a binary string is a string consisting of symbols $$$0$$$ and $$$1$$$). In an operation, you select a prefix of $$$a$$$, and simultaneously invert the bits in the prefix ($$$0$$$ changes to $$$1$$$ and $$$1$$$ changes to $$$0$$$) and reverse the order of the bits in the prefix.

For example, if $$$a=001011$$$ and you select the prefix of length $$$3$$$, it becomes $$$011011$$$. Then if you select the entire string, it becomes $$$001001$$$.

Your task is to transform the string $$$a$$$ into $$$b$$$ in at most $$$3n$$$ operations. It can be proved that it is always possible.

Input format:
The first line contains a single integer $$$t$$$ ($$$1\le t\le 1000$$$)  — the number of test cases. Next $$$3t$$$ lines contain descriptions of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1\le n\le 1000$$$)  — the length of the binary strings.

The next two lines contain two binary strings $$$a$$$ and $$$b$$$ of length $$$n$$$.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$1000$$$.

Output format:
For each test case, output an integer $$$k$$$ ($$$0\le k\le 3n$$$), followed by $$$k$$$ integers $$$p_1,\ldots,p_k$$$ ($$$1\le p_i\le n$$$). Here $$$k$$$ is the number of operations you use and $$$p_i$$$ is the length of the prefix you flip in the $$$i$$$-th operation.

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

Note:
In the first test case, we have $$$01\to 11\to 00\to 10$$$.

In the second test case, we have $$$01011\to 00101\to 11101\to 01000\to 10100\to 00100\to 11100$$$.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length $$$2$$$, which will leave $$$a$$$ unchanged.

Example input:
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1

Example output:
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Editorial:
The easy version has two main solutions:
Solution 1: $$$O(n)$$$ time with $$$3n$$$ operations
The idea is to fix the bits one-by-one. That is, make $$$s_1=t_1$$$, then make $$$s_2=t_2$$$, etc. To fix the bit $$$i$$$ (when $$$s_i\ne t_i$$$), we can flip the prefix of length $$$i$$$, then flip the prefix of length $$$1$$$, and again flip the prefix of length $$$i$$$. These three operations do not change any other bits in $$$s$$$, so it's simple to implement in $$$O(n)$$$. Since we use $$$3$$$ operations per bit, we use at most $$$3n$$$ operations overall.
Solution 2: $$$O(n^2)$$$ time with $$$2n$$$ operations
In this solution, we take a similar approach to solution 1, in that we fix the bits one-by-one. This time, we will fix the bits in reverse order. To fix the bit $$$i$$$, we can either flip the prefix of length $$$i$$$, or flip the first bit and then flip the prefix of length $$$i$$$. Since we do this in reverse order, the previously fixed bits do not get messed up by this procedure. And we use at most $$$2$$$ operations per bit, so $$$2n$$$ operations overall.
However, we do have to simulate the operations in order to check if we should flip the first bit. Simulating an operation can easily be done in $$$O(n)$$$ time per operation, or $$$O(n^2)$$$ time to simulate all operations.