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:
from math import ceil
for i in range(int(input())):
    n = int(input())
    ss = input()
    sss = input()
    s1 = []
    s2 = []
    for j in ss:
        s1.append(int(j))
    for j in sss:
        s2.append(int(j))
    ans1 = 0
    ans2 = []
    for j in range(len(s1) - 1, -1, -1):
        if s1[j] != s2[j]:
            if s1[0] == s1[j]:
                ans1 += 1
                ans2.append(j + 1)
                if (j + 1) % 2 == 1:
                    s1[(j + 1) // 2] += 1
                    s1[(j + 1) // 2] %= 2
                    for k in range((j + 1) // 2):
                        s1[k], s1[2 * ((j + 1) // 2) - k] = (s1[2 * ((j + 1) // 2) - k] + 1) % 2, (s1[k] + 1) % 2
                else:
                    for k in range((j + 1) // 2):
                        s1[k], s1[2 * ((j + 1) // 2) - k - 1] = (s1[2 * ((j + 1) // 2) - 1 - k] + 1) % 2, (s1[k] + 1) % 2
            else:
                ans1 += 2
                ans2.append(1)
                s1[0] = (s1[0] + 1) % 2
                ans2.append(j + 1)
                if (j + 1) % 2 == 1:
                    s1[(j + 1) // 2] += 1
                    s1[(j + 1) // 2] %= 2
                    for k in range((j + 1) // 2):
                        s1[k], s1[2 * ((j + 1) // 2) - k] = (s1[2 * ((j + 1) // 2) - k] + 1) % 2, (s1[k] + 1) % 2
                else:
                    for k in range((j + 1) // 2):
                        s1[k], s1[2 * ((j + 1) // 2) - 1 - k] = (s1[2 * ((j + 1) // 2) - 1 - k] + 1) % 2, (s1[k] + 1) % 2
    print(ans1, end = ' ')
    print(*ans2)