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.*;
import java.io.*;
 
public class Main {
 
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        w:while (t-->0){
            int n = sc.nextInt();
            char[]a = sc.next().toCharArray();
            char[]b = sc.next().toCharArray();
            ArrayList<Integer>ans = new ArrayList<>(),zeroids=new ArrayList<>();
            for (int i =0;i<n;i++){
                zeroids.add(i);
                zeroids.add(n-i-1);
            }
            int id=0;
            long cur =0;
            for (int i =n-1;i>=0;i--){
                int zeroid=zeroids.get(id++);
                int flip=(n-i-1)%2;
                char v = flip==1?a[zeroid]=='1'?'0':'1':a[zeroid];
                if (v!=b[i]){
                    ans.add(i+1);
                }
                else {
                    ans.add(1);
                    ans.add(i+1);
                }
                cur=i-cur;
            }
            pw.print(ans.size());
            for (int x:ans)
                pw.print(" "+x);
            pw.println();
        }
 
        pw.close();
 
    }
    static class Scanner {
        BufferedReader br;
        StringTokenizer st;
 
        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }
 
        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
 
        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
 
        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }
 
        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
 
        public int[] nextIntArr(int n) throws IOException {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = Integer.parseInt(next());
            }
            return arr;
        }
 
    }
 
}