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.ArrayList;
import java.util.Scanner;
import java.util.Vector;

public class prifix_flip_easy {
//    public static String reverse(String str,int n){
//        String str1="";
//        String str2=str.substring(0,n);
//        //System.out.println(str2);
//        for(int i=str2.length()-1;i>=0;i--){
//            if(str2.charAt(i)=='1') str1+="0";
//            else str1+="1";
//        }
//        //System.out.println(str1);
//        return str1+str.substring(n);
//    }
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        //System.out.println(reverse("01011",5));
        int t = input.nextInt();
        while(t!=0){
            t--;
            int n = input.nextInt();
            String a =input.next();
            a+="0";
            String b = input.next();
            b+="0";
            //ArrayList<Integer> array = new ArrayList<>();
            Vector<Integer> array = new Vector<>();
            for(int i =0;i<n;i++){
                if(a.charAt(i)!=a.charAt(i+1)){
                    array.add(i+1);
                }
                //System.out.print(a.charAt(i)+" ");
            }
            //System.out.println();
            for(int i=n-1;i>=0;i--){
                if(b.charAt(i)!=b.charAt(i+1)){
                    array.add(i+1);
                }
            }
            System.out.print(array.size()+" ");
            for(int i =0;i<array.size();i++){
                System.out.print(array.get(i)+" ");
            }
        }
    }
}
