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.
__MULTITEST = True

## solve

def solve():
    n = int(input())
    a = [int(d) for d in str(input())]
    b = [int(d) for d in str(input())]

    order = []
    inver = []
    p1 = 0; p2 = n-1;
    turn = 0;
    while p1 <= p2:
        if turn == 0:
            order.append(p1)
            inver.append(turn)
            p1 += 1; turn ^= 1;
        else:
            order.append(p2)
            inver.append(turn)
            p2 -= 1; turn ^= 1;

    op = []
    i = 0
    while i < n:
        if a[order[i]] ^ inver[i] == b[n - i - 1]:
            op.append(1)
            a[order[i]] ^= 1
        else:
            op.append(n - i)
            i += 1

    #print(a)
    #print(b)

    print(len(op), end = ' ')
    for x in op:
        print(x, end = ' ')
    print('')


## main
if __name__ == "__main__":
    t = (int(input()) if __MULTITEST else 1)
    for tt in range(t):
        solve();