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.
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.
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.
for t in range(int(input())):
    n = int(input())
    a = input()
    b = input()
    ans = []
    for i in range(n - 1, -1, -1):
        if a[i] != b[i]:
            ans.append(i + 1)
            ans.append(1)
            ans.append(i + 1)
    if ans:
        print(len(ans), *ans)
    else:
        print(0)
for t in range(int(input())):
    n = int(input())
    a = input()
    b = input()
    ans = []
    for i in range(n - 1, -1, -1):
        if a[i] != b[i]:
            ans.append(i + 1)
            ans.append(1)
            ans.append(i + 1)
    if ans:
        print(len(ans), *ans)
    else:
        print(0)
t = int(input())
while t > 0:
    n = int(input())
    a = input()
    b = input()
    j = 0
    l = 0
    ans = []
    for i in range(0, n):
        if a[i] == b[j]:
            j += 1
            continue
        else:
            l += 1
            ans.append(j)
            if j > 0:
                l += 1
                ans.append(0)
                l += 1
                ans.append(j)
            j += 1
    t = t - 1
    print(l)
    if len(ans) > 0:
        for i in ans:
            print(i + 1, end=' ')
        print()