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 java.util.*;
public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int a,b,c,d;
    int e,f,g,h,i,j;
    i = sc.nextInt();
    for(j = 0;j < i;j++){
    a = sc.nextInt();
    String A = sc.next();
    String B = sc.next();
    b = 0;
    String C = "";
    char m,n,o,p;
    for(c = (a-1);c >= 1;c--){
      String F = A.substring(0,1);
      m = A.charAt(0);
      n = B.charAt(c);
      //System.out.println(m+" "+n);
      if(m == n){
        if(c == 0){
          break;
        }
        C = C + "1 ";
        b++;
        if(m == '0'){
          m = '1';
          F = "1";
        }
        else{
          m = '0';
          F = "0";
        }
        
      }
      //String F = Character.toString(m);
      String E =  F + A.substring(1,(c+1));
      //System.out.println(E+"c"+F);
      String D = "";
      for(e = c;e >= 0 ; e--){
        if(E.charAt(e) == '0'){
          D = D + "1";
        }
        else{
          D = D + "0";
        }
      }
      A = D + A.substring((c+1),a);
      //System.out.println(D+"a");
      //System.out.println(A+"b");
      C = C + (c+1) + " ";
      b++;
    }
    m= A.charAt(0);
    n = B.charAt(0);
    //System.out.println(C);
    if(m != n){
      b++;
      C = C + "1 ";
    }
    //System.out.println(A+" "+B);
    if(b != 0){
      C = C.substring(0,C.length()-1);
    }
    if(b == 0){
      System.out.println(b);
    }
    else{
      System.out.println(b+" "+C);
    }
    }
  }
}