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.util.Scanner;
import java.awt.image.BandedSampleModel;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class C {


    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int cases = sc.nextInt();
        for(int i=0;i<cases;i++)
        {
            int n = sc.nextInt();
            int arr1[] = new int[n];
            int arr2[]=new int[n];
            String s1=sc.next();
            String s2=sc.next();
            for(int j=0;j<n;j++)
            {
                arr1[j]=s1.charAt(j)-'0';
            }
            for(int j=0;j<n;j++)
            {
                arr2[j]=s2.charAt(j)-'0';
            }

            StringBuilder ans=new StringBuilder();
            int count=0;
            boolean flip=false;
            for(int j=0;j<n;j++)
            {
                //if((arr1[j]!=arr2[j]&&!flip)||(flip&&arr1[j]==arr2[j]))
                if(arr1[j]!=arr2[j])
                {

                    if(j==0)
                    {
                        ans.append(1+" ");
                        count++;
                    }
                    else if(j!=n-1)
                    {
                        ans.append((j+1)+" "+1+" "+(j+1)+" ");
                        count=count+3;
                        flip=!flip;
                    }
                    else
                    {
                        ans.append(n+" "+1+" "+n+" ");
                        count=count+3;
                    }
                }
            }
            System.out.println(count+" "+ ans);


        }

    }
}