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 io
import os

from collections import Counter, defaultdict, deque


def solve(N, A, B):
    #print()
    #print()
    def f(A):
        p = []
        front = []
        i = 0
        #print('A', A)
        while i < N:
            x = A[i]
            if not front:
                front.append([x, 1])
            elif front[-1][0] != x:
                if len(front) == 1:
                    front.append([x, 1])
                else:
                    assert len(front) == 2
                    #print('flip', front[0][1])
                    p.append(front[0][1])
                    front[0][0] ^= 1
                    front[0][1] = front[0][1] + front[1][1]
                    front[1][0] = x
                    front[1][1] = 1
                    assert front[0][0] != front[1][0]
            else:
                assert front[-1][0] == x
                front[-1][1] += 1
            i += 1
            #print(*[[y] * c for y, c in front], A[i:])
        if len(front) == 2:
            #print('flip', front[0][1])
            p.append(front[0][1])
            front[0][0] ^= 1
        #print('parity', front[0][0])
        return front[0][0], p

    parity1, p1 = f(A)
    parity2, p2 = f(B)
    if parity1 != parity2:
        p1.append(N)
    p2 = deque(p2)
    while p1 and p2 and p1[-1] == p2[-1]:
        p1.pop()
        p2.pop()
    p2 = list(p2)[::-1] 
    #assert len(p1) + len(p2) <= 2 * N
    return str(len(p1) + len(p2)) + ' ' + ' '.join([str(x) for x in p1 + p2])


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    T = int(input())
    for t in range(T):
        N, = [int(x) for x in input().split()]
        A = list(map(int, input().decode().rstrip()))
        B = list(map(int, input().decode().rstrip()))
        ans = solve(N, A, B)
        print(ans)

