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 JAVA):

Solution:
import java.util.*; 
public class C_658{
public static void main(String[] args)
{
   Scanner omkar = new Scanner(System.in);
   int cases = omkar.nextInt();
   int a, b, n;
   int[] arr;
   int[] arr2;
   int[] ret;
   int pos;
   String st;
   for(int k = 0; k < cases; k++)
   {
      a = 0;
      b = 1;
      pos = 0;
      n = omkar.nextInt();
      ret = new int[3*n];
      arr = new int[n];
      arr2 = new int[n];
      st = omkar.next();
      for(int i = 0; i < n; i++)
      {
         arr[i] = st.charAt(i) - 48;
      }
      st = omkar.next();
      for(int i = 0; i < n; i++)
      {
         arr2[i] = st.charAt(i) - 48;
      }
      for(int i = n-1; i >= 0; i--)
      {
         if(((arr[a+b*i]^(Math.max(-1*b, 0))) ^ arr2[i]) == 1)
         {
            if(((arr[a]^(Math.max(-1*b, 0))) ^ arr2[i]) == 0)
            {
              ret[pos] = 1;
              pos++;   
            }
            ret[pos] = i+1;
            pos++; 
            a = a+b*(i);
            b = -1*b;
         }
      }
      System.out.print(pos + " ");
      for(int i = 0; i < pos; i++)
      {
         System.out.print(ret[i] + " ");
      }
      System.out.println("");
   }
      }
}