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 PYTHON):

Solution:
T = input()
res = []
for t in range(T) :
  N = input()
  a = raw_input()
  b = raw_input()

  L, R = 0, N-1
  front = a[0]
  flip = False
  ans = []

  #print ""
  #print "A:", a
  #print "B:", b

  for n in range(N-1,-1,-1) :
    curA = int(a[R]) if not flip else 1-int(a[L]) 

    #print "A B:", curA, b[n]
    #print "Front:", front
    #print "L R:", L, R

    if str(curA) == b[n] : 
      if flip : L += 1
      else : R -= 1
      continue

    if str(front) == b[n] : ans.append(1)
    ans.append(R-L+1)
    flip = not flip
    front = 1-curA

    if flip : L += 1
    else : R -= 1

    #print ans
    #print ""

  res.append("%d %s" % (len(ans), ' '.join(str(x) for x in ans)))

print '\n'.join(res)