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.
def flipsub(L, to_index):
    for i in range((to_index + 1)//2):
        f = L[i]
        s = L[to_index - i]
        ff = '0' if f  == '1' else '1'
        fs = '0' if s == '1' else '1'
        L[i] = fs
        L[to_index-i] = ff
        #print('Swapping index {} left {} rihgt {}'.format(i, fs, ff))
    if to_index % 2 == 0:
        #swap middle element
        mid = to_index //2
        L[mid] = '0' if L[mid] == '1' else '1'

def flip(A, B):
    out = []
    for index in range(N-1, -1, -1):
        a = A[index]
        b = B[index]
        if a != b:
            first = A[0]
            if first == b:
                #first flip first
                flipsub(A, 0)
                out.append(1)

            flipsub(A, index)
            out.append(index + 1)
    return out

def test(flips, A, B):
    print('TESTING ****************')
    print(A)
    for flip in flips:

        flipsub(A, flip-1)
        print('Flipping ', flip)
        print(A)
    ok = True
    for i in range(len(A)):
        if A[i] != B[i]:
            ok = False
    if ok:
        print('PASSED')
    else:
        print('FAILED')

def test2():
    A = list('11')
    # 0010011001
    flipsub(A, 1)
    print(A)

#test2()
#exit()

T = int(raw_input())
for _ in range(T):
    N = int(raw_input())
    s = raw_input()
    A = list(s)
    C = list(s)
    B = list(raw_input())
    out = flip(A, B)
    s = ' '.join(map(str, out))
    print('{} {}'.format(len(out), s))
    #test(out, C, B)