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 sys

def invertflip(lst):
    lst.reverse()
    for i in range(len(lst)):
        if lst[i] == '1':
            lst[i] = '0'
        else:
            lst[i] = '1'

    return lst

def answer(n, a, b):
    steps = 0
    rptr = n-1
    len_ary = []
    a = list(a) #for reversing
    b = list(b)
    #print('a=', a, 'b=', b)
    while rptr >= 0:
        #print('rptr=', rptr)
        if a[rptr] == b[rptr]:
            #print('last element same')
            rptr -= 1
            continue
        # not same.
        if a[0] != b[rptr]: #ideal
            #invert and flip entire substring
            #print('inverting and flipping', a[0:rptr+1])
            a[0:rptr+1] = invertflip(a[0:rptr+1])
            #print('now', a[0:rptr+1])
            len_ary.append(rptr+1)
            steps += 1
            rptr -= 1
            continue
        # a[0] is not what I want, and I use a step to change it.
        #invert and flip 1st element.
        #print('inverting and flipping', a[0:1])
        a[0:1] = invertflip(a[0:1])
        #print('now', a[0:1])
        len_ary.append(1)
        steps += 1
        #rptr does not change.  Will do the #ideal one next go around.
    #print('steps=', steps, 'len_ary=', len_ary)
    ans = [steps]
    ans.extend(len_ary)
    return ans #list of ints

def main():
    t = int(input())
    while t:
        n = int(sys.stdin.readline().rstrip())
        a = sys.stdin.readline().rstrip()
        b = sys.stdin.readline().rstrip()
        print(' '.join(map(str, answer(n, a, b))))

        t -= 1
    return
main()