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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.util.*;

public class vk18
{
    public static void main(String[] stp) throws Exception
    {
        Scanner scan = new Scanner(System.in);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        //PrintWriter printWriter = new PrintWriter(System.out);
        StringTokenizer st = new StringTokenizer(br.readLine());
        int t=Integer.parseInt(st.nextToken());
        while (t-- > 0)
        {
            st = new StringTokenizer(br.readLine());
            int n=Integer.parseInt(st.nextToken()),i=0,count = 0,q=0;
            st = new StringTokenizer(br.readLine());
            String str1 = st.nextToken();
            char []c1 = str1.toCharArray();
            st = new StringTokenizer(br.readLine());
            String str2 = st.nextToken();
            Integer ans [] = new Integer[3*n];
            char []c2 = str2.toCharArray();
            for(i=n-1;i>=0;i--)
            {
                if(c1[i] == c2[i]) continue;
                if(c1[i] == c1[0])
                {
                    ans[q++] = i+1;
                    //pw.print(i+" ");
                    count++;
                }
                else
                {
                    ans[q++] = 1;
                    ans[q++] = i+1;
                    count+=2;
                    c1[0] = c1[i];
                }
                for(int j = 0 ; j<=i/2 ;j++)
                {
                    if(c1[j] == '0') c1[j] = '1';
                    else { c1[j] = '0'; }
                    if(j != i-j)
                    {
                        if(c1[i-j] == '0') c1[i-j] = '1';
                        else { c1[i-j]= '0'; }
                    }
                    char na = c1[j];
                    char ch = c1[i-j];
                    c1[j] = ch;
                    c1[i-j] = na;


                }



            }

            pw.print(count+" ");
            for(i=0;i<q;i++) pw.print(ans[i]+" ");
            pw.println();
            pw.flush();
        }

    }
}