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:
import sys
input = sys.stdin.readline

T = int(input())

# def check(A, B, ops):
#     for op in ops:
#         # print('op', op)
#         # print('A', A)
#         A = ['1' if el == '0' else '0' for el in A[:op]][::-1] + A[op:]
#         # print('A after', A)

#     if A != B:
#         print(A, B, ops)
#         assert False

for t in range(T):
    N = int(input())

    A = input()[:-1]
    B = input()[:-1]
    # simuA = list(A)

    answer = []

    el0 = A[0]
    for i in range(N-1, -1, -1):
        # print('i', i)
        # print('A is', simuA)
        # print('el0', el0)
        if el0 == B[i]:
            answer.append(1)
            # simuA[0] = '0' if simuA[0] == '1' else '1'
            # print('B[i] is', B[i])
            # print('simuA is', simuA)
        answer.append(i+1)
        # simuA = ['1' if el == '0' else '0' for el in simuA[:i+1]][::-1] + simuA[i+1:]
        # print('simuA is', simuA)
        if i == N-1:
            el0 = '0' if A[i] == '1' else '1'
        elif (N-i) % 2:
            el0 = '0' if A[N-1-(N-i-1)//2] == '1' else '1'
        else:
            # print('Setting el0 because pair', N-i, A[(N-i)//2])
            el0 = A[(N-i)//2]

    print(len(answer), ' '.join(map(str, answer)))

    # check(list(A), list(B), answer)