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:



list_input = lambda: list(map(int, input().split()))
int_input = lambda: int(input())

def count_vals(l):
    d = {}
    for i in l:
        d[i] = d.get(i, 0) + 1
    return d


def bit_inv(b):
    if b == '1':
        return '0'
    else:
        return '1'

def opp(s, n):
    for i in range((n + 1) // 2):
        # print(s[i], s[n-i])
        s[i], s[n-i] = bit_inv(s[n-i]), bit_inv(s[i])
        
    if (n + 1) % 2:
        s[(n + 1) // 2] = bit_inv(s[(n + 1) // 2])

    # print(n, ''.join(s))
    return s


def solve(a, b, n):
    l = []
    for i in range(n-1, 0, -1):
        x = a[0]
        if x == b[i]:
            a[0] = bit_inv(a[0])
            l.append(1)
            l.append(i+1)
        else:
            l.append(i+1)

        # a[0] = bit_inv(a[i])
        # print(i, a[0])
        a = opp(a, i)
        # print(a)


    if a[0] != b[0]:
        l.append(1)
        a[0] = bit_inv(a[0])

    return l

#
#
# for _ in range(int_input()):
#     n = int_input()
#     a = list(input())
#     b = list(input())
#
#     l = []
#     for i in range(n-1, 0, -1):
#         x = a[0]
#         if x == b[i]:
#             l.append(1)
#             l.append(i+1)
#         else:
#             l.append(i+1)
#         a[0] = bit_inv(a[i])
#         print(i, a[0])
#
#
#     if a[0] != b[0]:
#         l.append(1)
#         a[0] = bit_inv(a[0])
#
#     print(len(l), end=" ")
#     print(*l)


for _ in range(int(input())):
    n = int_input()
    a = list(input())
    b = list(input())

    l = solve(a, b, n)

    print(len(l), end=" ")
    print(*l)