Problem: 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

The first line contains a single integer t (1≤ t≤ 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≤ n≤ 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

For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ 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.

Example

Input


5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1


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

Note

In the first test case, we have 01→ 11→ 00→ 10.

In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 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.

Solution (in PYTHON3):

Solution:
T = int(input())
def go(x, k):
    ans = []
    for i in range(k+1):
        ans.append(x[k-i]^1)
    return ans

for t in range(T):
    n = int(input())
    source = [int(i) for i in input()]
    target = [int(i) for i in input()]
    left = 0
    right = n-1
    ans = []
    direction = -1
    flip_counter = 0
    pointer = n-1
    while pointer > 0:
        if direction < 0:
            source_pointer = right
            oppo = left
        else:
            source_pointer = left 
            oppo = right
        if source[source_pointer] ^flip_counter != target[pointer]: # flip is required
            # check if the first bit of source is the same as the target upon flipping
            if source[oppo] ^ flip_counter == target[pointer]:
                ans.append(1)
                # source[0] ^= 1
            ans.append(pointer + 1)
            flip_counter ^= 1
            # don't do this: source = go(source, pointer)
            direction *= -1
        if direction < 0:
            right -= 1
        else:
            left += 1
        pointer -= 1
    # now pointer is at 0, just check the first bit
    if direction < 0:
        source_pointer = right
    else:
        source_pointer = left 
    if source[source_pointer] ^ flip_counter != target[0]:
        ans.append(1)
    m = len(ans)
    ans = [m] + ans
    print(*ans)