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.
// Don't place your source in a package
import java.lang.reflect.Array;
import java.text.DecimalFormat;
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
import java.util.stream.Stream;


// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    static class FastScanner {//scanner from SecondThread
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int Int() {
            return Integer.parseInt(next());
        }

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

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(System.out);



        int T=Int();
        for(int t=0;t<T;t++){
            int n=Int();
            String a=Str();
            String b=Str();

            Solution sol=new Solution();
            sol.solution(out,a,b);
        }
        out.flush();

    }

    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}



class Solution{
    public void solution(PrintWriter out,String s,String t){
        int A[]=new int[s.length()];
        int B[]=new int[t.length()];
        for(int i=0;i<A.length;i++){
            A[i]=s.charAt(i)-'0';
            B[i]=t.charAt(i)-'0';
        }


        int j=B.length-1;
        int l=0,r=s.length()-1;
        int cnt=0;
        List<Integer>res=new ArrayList<>();

        while(j>=0){
            if(j==0){
                if((A[l]+cnt)%2!=B[0]){
                    res.add(0);
                }
                break;
            }

            if(cnt%2==0){
                if(A[r]==B[j]){//just skip
                    r--;
                }
                else{
                    if((A[l]+1)%2==B[j]){
                        res.add(j);
                    }
                    else{
                        res.add(0);
                        res.add(j);
                    }
                    l++;
                    cnt++;
                }

            }
            else{
                if((A[l]+cnt)%2==B[j]){
                    l++;
                }
                else{
                    if((A[r])%2==B[j]){
                        res.add(j);
                    }
                    else{
                        res.add(0);
                        res.add(j);
                    }
                    r--;
                    cnt++;
                }
            }
            j--;
        }




        if(res.size()==0){
            out.println(res.size()+" ");
            return;
        }
        out.print(res.size()+" ");
        for(int i:res){
            out.print((i+1)+" ");
        }
        out.println();

    }


}


