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.*;

public class Main {

    public static Scanner sc = new Scanner(System.in);
    public static int inf=1000000007;

    public static void main(String[] args) {

        long t=sc.nextLong();
        while(t-->0){

            int n=sc.nextInt();

            String a=sc.next();
            String b=sc.next();

            ArrayList<Integer> al= new ArrayList<>();

            for(int i=0;i<n;i++){
                if(a.charAt(i)!=b.charAt(i)){
                    al.add(i+1);
                    al.add(1);
                    al.add(i+1);
                }
            }

            System.out.print(al.size()+" ");
            al.forEach(x->System.out.print(x+" "));

            System.out.println();


            

        }

    }

    ///////For printing elements in array

    public static void print(int[] arr){
        for(int x:arr){
            System.out.print(x+" ");
        }
        System.out.println();
    }

    ///////For freq of elements in array

    public static int[] freq(int[] freq_arr,int[] arr){
        for(int i=0;i<arr.length;i++){
            freq_arr[arr[i]]++;
        }

        return freq_arr;
    }


    ///////For sum elements in array

    public static int sum(int[] arr){

        int sum=0;

        for(int i=0;i<arr.length;i++){
            sum+=arr[i];
        }
        return sum;
    }

///////For storing elements in array

    public static int[] scan(int[] arr){
        for(int i=0;i<arr.length;i++){
            arr[i]=sc.nextInt();
        }

        return arr;
    }

///////For GCD

    public static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

////////Forprime

    public static boolean isPrime(long n)
    {
        if(n < 2) return false;
        if(n == 2 || n == 3) return true;
        if(n%2 == 0 || n%3 == 0) return false;
        long sqrtN = (long)Math.sqrt(n)+1;
        for(long i = 6L; i <= sqrtN; i += 6) {
            if(n%(i-1) == 0 || n%(i+1) == 0) return false;
        }
        return true;
    }




}