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 A{
    static PrintWriter out;
    static InputReader in;
    public static void main(String args[]){
        out = new PrintWriter(System.out);
        in = new InputReader();
        new A();
        out.flush(); out.close();
    }   
    A(){
        solve();
    }
    int p;
    ArrayDeque<Integer> ad = new ArrayDeque<>();
    void remove(){
        if(p == 0){
            ad.pollLast();
        }else{
            ad.pollFirst();
        }
    }
    void change(){
        if(p == 0){
            ad.addFirst(1 - ad.pollFirst());
        }else{
            ad.addLast(1 - ad.pollLast());
        }
    }
    void solve(){
        int t = in.nextInt();
        o : while(t-- > 0){
            int n = in.nextInt();
            char a[] = in.next().trim().toCharArray();
            char b[] = in.next().trim().toCharArray();
            ad.clear();
            ArrayList<Integer> al = new ArrayList<>();
            boolean rev = false; p = 0;
            for(int i = 0; i < n; i++)ad.addLast(a[i] - '0');
            for(int i = n - 1; i >= 0; i--){
                int vals = -1, vale = -1;
                if(p == 0){
                    vals = ad.getFirst();
                    vale = ad.getLast();
                }else{
                    vale = ad.getFirst();
                    vals = ad.getLast();
                }
                if(p == 1){
                    vals = 1 - vals; vale = 1 - vale;
                } 
                // System.out.println(i + " " + p + " " + rev + " " + vals + " " + vale + " " + ad);
                if(vale == b[i] - '0'){
                    remove();
                }else{
                    if(vals == vale){
                        al.add(ad.size());
                        rev = !rev; p = 1 - p;
                        if(vals == b[0] - '0'){
                            change();
                            al.add(1);
                        }
                    }else{
                        change();
                        al.add(1); 
                        al.add(ad.size());
                        rev = !rev; p = 1 - p;
                    }
                    remove();
                }
            }
            out.print(al.size() + " ");
            for(int v : al)out.print(v + " ");
            out.println();
        }
    }
    public static class InputReader{
        BufferedReader br;
        StringTokenizer st;
        InputReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public int nextInt(){
            return Integer.parseInt(next());
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
        public String next(){
            while(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch(IOException e){}
            }
            return st.nextToken();
        }
    }
}
        