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.*;
import java.util.*;

public final class codeForces
{
    static String rev(String X)
    {
        String Y="";
        for(int i=X.length()-1; i>=0; i--)Y+=(X.charAt(i)=='1')?"0":"1";
        return Y;
    }

    

    public static void main(String args[])throws IOException
    {
        BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        // Scanner in=new Scanner(System.in);
        //Integer.parseInt(in.readLine());
        //Integer.parseInt(st.nextToken());
        //Long.parseLong(in.readLine());
        //Long.parseLong(st.nextToken());

        {
            
           int T=Integer.parseInt(in.readLine());
            while(T-->0)
            {
                int N=Integer.parseInt(in.readLine());
                String A=in.readLine();

                // System.out.println(A);
                //System.out.println(A1);       
                String print="";
                String B=in.readLine();
                int start=0,end=N-1;
                int count=0;

                for(int i=N-1; i>=0; i--)
                {
                    if(B.charAt(i)!=A.charAt(i))
                    {
                        if(A.charAt(0)==B.charAt(i))
                        {
                            print+=" 1";
                            A=((A.charAt(0)=='0')?"1":"0")+A.substring(1);
                            
                            count+=1;
                        }
                        if(i>0)
                        {                            
                            A=rev(A.substring(0,i+1));                            
                           print+=" "+(i+1);
                            count+=1;
                        }
                    }
                }
                if(A.charAt(0)!=B.charAt(0)){
                count++; print+=" 1";
                }

                System.out.println(count+" "+print);
                
            } 
        }
    }
}

/* StringTokenizer st;
st=new StringTokenizer(in.readLine());
int n=Integer.parseInt(st.nextToken());
int q=Integer.parseInt(st.nextToken());
int c=Integer.parseInt(st.nextToken());
int A[][]=new int[n][3];
for(int i=0; i<n; i++)
{
st=new StringTokenizer(in.readLine());
A[i][0]=Integer.parseInt(st.nextToken());
A[i][1]=Integer.parseInt(st.nextToken());
A[i][2]=Integer.parseInt(st.nextToken());
}
while(q-->0)            
{
st=new StringTokenizer(in.readLine());
long t=Long.parseLong(st.nextToken());
int x1=Integer.parseInt(st.nextToken());
int y1=Integer.parseInt(st.nextToken());
int x2=Integer.parseInt(st.nextToken());
int y2=Integer.parseInt(st.nextToken());
long b=0;
for(int i=0; i<n; i++)
{
int x=A[i][0];
int y=A[i][1];

if(x>=x1 && x<=x2 && y>=y1 && y<=y2)
{
b+=(A[i][2]+t)%(c+1);
}
}
System.out.println(b);
}            
 */