CF1381A1.Prefix Flip (Easy Version)

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.

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 $1000$ .

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.

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.