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.
# input = open('file.txt').readline

def chng(c):

  if c == '1':
    return '0'
  else:
    return '1'

def ulta(ss , i):

  

  frt = ss[:i+1]
  bck = ss[i+1:]

  frt = list(frt[::-1])

  for j in range( len(frt) ):

    frt[j] = chng( frt[j] )
  
  frt = ''.join(frt)
  # print('ulta of', ss , i, 'to', frt+bck)
  return frt + bck

for _ in range( int(input()) ):

  n = int(input().strip())
  sa = input().strip()
  sb = input().strip()

  if sa == sb:
    print(0)
    continue

  ft = sa[0]

  i = n
  ans = 0
  arr = []

  while i > 0:
    i -= 1

    # print(sa)

    if sa[i] == sb[i]:
      continue

    elif ft != sb[i]:

      ans += 1
      arr.append(i+1)

      sa = ulta( sa , i )
      ft = sa[0]
    
    elif ft == sb[i] and i == 0:
      break

    else:

      ans += 1
      arr.append(1)

      sa = ulta( sa , 0 )

      if i != 0:
        ans += 1
        arr.append(i+1)

      sa = ulta( sa , i )
      ft = sa[i]


  
  print( ans , *arr )


