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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Prefix_F {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
    public static void main(String[] args) {
         FastReader in = new FastReader();
         int t = in.nextInt();
         StringBuilder sb = new StringBuilder();
         while(t-- > 0){
             int n = in.nextInt();
             String a = in.nextLine();
             String b = in.nextLine();
             char arr[] = a.toCharArray();
             char brr[] = b.toCharArray();
             int count = 0;
             String ans = "";
             for(int i = n - 1; i >= 0; i--){
                 if(arr[i] != brr[i]){
                     if(brr[i] == arr[0]){
                         count+=2;
                         ans+="1 ";
                         if(arr[0] == '1'){
                             arr[0] = '0';
                         }else{
                             arr[0] = '1';
                         }
                         arr = fun(arr,i);
                         int def = i + 1;
                         ans+=def;
                         ans+=" ";
                     }else{
                         count++;
                         arr = fun(arr,i);
                         int def = i + 1;
                         ans+=def;
                         ans+=" ";
                     }
                 }
             }

         /*    for(int i = 0; i < arr.length; i++){
                 System.out.print(arr[i]+" ");
             }
             System.out.println();

          */

             sb.append(count+" "+ans+"\n");


         }
         System.out.println(sb);
    }

    static char [] fun(char arr[], int pre){
        char res[] = new char[arr.length];
        for(int i = 0; i <= pre; i++){
            if(arr[i] == '1'){
                arr[i] = '0';
            }else{
                arr[i] = '1';
            }
        }
        int k = 0;
        for(int i = pre; i >= 0; i--){
            res[k] = arr[i];
            k++;
        }
        for(int i = pre + 1; i < arr.length; i++){
            res[i] = arr[i];
        }

        return res;
    }
}