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.
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s); sys.stdout.write('\n')
def wi(n): sys.stdout.write(str(n)); sys.stdout.write('\n')
def wia(a, sep=' '): sys.stdout.write(sep.join([str(x) for x in a])); sys.stdout.write('\n')


# def solve(n, a, b):
#     res = []
#     for i in range(n-1, -1, -1):
#         if a[i] == b[i]:
#             continue
#         if a[0] == b[i]:
#             a[0] = 1 - a[0]
#             res.append(1)
#
#         if i % 2 == 0:
#             a[i // 2] = 1 - a[i // 2]
#         for j in range((i + 1) // 2):
#             a[j], a[i - j] = 1 - a[i - j], 1 - a[j]
#         res.append(i + 1)
#
#     return [len(res)] + res

def solve(n, a, b):
    a.append(0)
    b.append(0)

    res1 = []
    res2 = []
    for i in range(n):
        if a[i] != a[i+1]:
            res1.append(i+1)
        if b[i] != b[i+1]:
            res2.append(i+1)

    res = res1 + res2[::-1]

    return [len(res)] + res


def main():
    for _ in range(ri()):
        n = ri()
        a = [int(si) for si in rs()]
        b = [int(si) for si in rs()]
        wia(solve(n, a, b))


if __name__ == '__main__':
    main()
