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 collections import deque


def solve(n, a, b):
    ans = deque()
    for i in range(n):
        if a[n - 1 - i] != b[n - 1 - i]:
            if a[0] == b[n - 1 - i]:
                # headflip
                # #print(1)
                ans.append(1)
                a[0] = 1 - a[0]
                # a[0:n-1-i] flip
                ##print(n - i)
                ans.append(n - i)
                # for j in range(n - i):
                #     a[j] = 1 - a[j]
                # a[:n - i + 1] = a[:n - i + 1:-1]
                st = deque([])
                for j in range(n - i):
                    st.append(1 - a[j])
                # print('before')
                # print(a)
                for j in range(n - i):
                    a[n - i - 1 - j] = st.popleft()
                # print('after')
                # print(a)
                # print(1)
                #print(n - i)
                # print(a)
            else:
                # a[0:n-1-i] flip
                ##print(n - i)
                ans.append(n - i)

                # for j in range(n - i):
                #     a[j] = 1 - a[j]
                # a[:n - i + 1] = a[:n - i + 1:-1]

                st = deque([])
                for j in range(n - i):
                    st.append(1 - a[j])

                # print('before')
                # print(a)
                for j in range(n - i):
                    a[n - i - 1 - j] = st.popleft()
                # print('after')
                # print(a)
                #print(n - i)
                # print(a)
    if len(ans) == 0:
        print(0)
        return
    else:
        print(len(ans), *ans)
        return


t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, list(input())))
    b = list(map(int, list(input())))
    solve(n, a, b)