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.
#
#    ------------------------------------------------
#           ____          _     Generatered using
#          / ___|        | |
#         | |    __ _  __| | ___ _ __  ______ _
#         | |   / _` |/ _` |/ _ \ '_ \|_  / _` |
#         | |__| (_| | (_| |  __/ | | |/ / (_| |
#          \____\____|\____|\___|_| |_/___\____|
#
#      GNU Affero General Public License v3.0
#    ------------------------------------------------
#    Author   : prophet
#    Created  : 2020-07-21 12:09:14.493283
#    UUID     : hmiUSX6K8LVodzyz
#    ------------------------------------------------
#
production = True

import sys, math, collections

def input(input_format = 0, multi = 0):

    if multi > 0: return [input(input_format) for i in range(multi)]
    else:
        next_line = sys.stdin.readline()[:-1]

        if input_format >= 10:
            use_list = False
            input_format = int(str(input_format)[-1])
        else: use_list = True

        if input_format == 0: formatted_input = [next_line]
        elif input_format == 1: formatted_input = list(map(int, next_line.split()))
        elif input_format == 2: formatted_input = list(map(float, next_line.split()))
        elif input_format == 3: formatted_input = list(next_line)
        elif input_format == 4: formatted_input = list(map(int, list(next_line)))
        elif input_format == 5: formatted_input = next_line.split()
        else: formatted_input = [next_line]

        return formatted_input if use_list else formatted_input[0]

def out(output_line, output_format = 0, newline = True):

    formatted_output = ""

    if output_format == 0: formatted_output = str(output_line)
    elif output_format == 1: formatted_output = " ".join(map(str, output_line))
    elif output_format == 2: formatted_output = "\n".join(map(str, output_line))

    print(formatted_output, end = "\n" if newline else "")

def log(*args):
    if not production:
        print("$$$", end = "")
        print(*args)

enu = enumerate

ter = lambda a, b, c: b if a else c

ceil = lambda a, b: -(-a // b)

def mapl(iterable, format = 0):
    
    if format == 0: return list(map(int, iterable))
    elif format == 1: return list(map(str, iterable))
    elif format == 2: return list(map(list, iterable))
#
#   >>>>>>>>>>>>>>> START OF SOLUTION <<<<<<<<<<<<<<
#

def f(a, l, r):
    
    k = []

    for i in a[l:r]:
        k.append(ter(i, 0, 1))

    #log("k", k)
    c = a[:l] + k[::-1] + a[r:]

    return c

def g(n):
    
    d = []

    t = 0
    #log(list(range(n//2)), list(range(n - 1, n//2 - 1, -1)))
    for i, j in zip(range(n//2 + 1), range(n - 1, n//2 - 1, -1)):
        t += 1
        d.append((i, t & 1))
        if len(d) == n:
            break
        t += 1
        d.append((j, t & 1))

    return d

def solve():

    n = input(11)
    a = input(4)
    b = input(4)


    s = []

    log(a)
#    for i, j in enu(b[::-1]):
#        if a[0] == j:
#            s.append(1)
#            a[0] = ter(j, 0, 1) 
#            log("flip")
#        
#        s.append(n - i)
#        #log(a, f(a, 0, n - i))
#        a = f(a, 0, n - i)
#        #log(n - i, a)
#
#    out([len(s)] + s, 1)

    d = g(n)

    log(d)

    dn = []

    for i, (j, k) in enu(d):
        #log(n - i - 1, b[n-i-1])
        dn.append((j, k, b[n - i - 1]))


    d = dn
    log(d)

    #d.sort()


    log(d)
    s2 = []

    if n == 1:
        if a == b:
            out(0)
        else:
            out("1 1")
        return
    log(a)

    for i, (j, k, bi) in enu(d):
        log(i, bi,  j, k)
        if k:
            k = ter(a[j], 0, 1)
        else:
            k = a[j]
        log(a[j], k)
        if bi != k:
            #log("f", i)
            s2.append(1)
        s2.append(n - i)

    out([len(s2)] + s2, 1)


    return


for i in range(input(11)): solve()
# solve()

#
#   >>>>>>>>>>>>>>>> END OF SOLUTION <<<<<<<<<<<<<<<
#
