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:
def f(n,a,b):
 a=map(int,a); b=map(int,b)
 out=[]; d=0
 while a:
  if d==0:
   if a[-1]==b[-1]: k=a.pop(); k=b.pop(); continue
   else:
    if len(a)>1:
     if a[0]==b[-1]:
      out.append(1); a[0]=1-a[0]
     out.append(len(a)); d=1-d
    else:
     if a[0]!=b[0]: out.append(1); a[0]=1-a[0]; k=a.pop(); k=b.pop()
  else:
   if (1-a[0])==b[-1]: k=a.pop(0); k=b.pop(); continue
   else:
    if len(a)>1:
     if (1-a[-1])==b[-1]:
      out.append(1); a[-1]=1-a[-1]
     out.append(len(a)); d=1-d
    else:
     if (1-a[-1])!=b[0]: out.append(1); a[-1]=1-a[-1]; k=a.pop(); k=b.pop()
 return [len(out)]+out

for t in xrange(int(raw_input())):
 n=int(raw_input())
 a=raw_input(); b=raw_input()
 print ' '.join(str(k) for k in f(n,a,b))