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:
/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
//package CodeforcesRound658;
import java.util.*;
/**
 *
 * @author Superuser
 */
public class PrefixFlipEasy {
    public static void main(String x[]) {
        Scanner t=new Scanner(System.in);
        int test=t.nextInt();
        while(test>0) {
            int n=t.nextInt();
            String a=t.next();
            String b=t.next();
            
            int count=0;
            ArrayList<Integer>prefix=new ArrayList<>();
            for(int i=n-1; i>=0; i--) {
                if(b.charAt(i)==a.charAt(i))
                    continue;
                else if(a.charAt(0)!=b.charAt(i)) {
                    String a1="";
                    for(int j=i; j>=0; j--) {
                        if(a.charAt(j)=='0')
                            a1+="1";
                        else
                            a1+="0";
                    }
                    for(int j=i+1; j<n; j++)
                        a1+=b.charAt(j);
                    a=a1;
                    prefix.add(i+1);
                    count++;
                } else {
                    String a1="";
                    for(int j=i; j>0; j--) {
                        if(a.charAt(j)=='0')
                            a1+="1";
                        else
                            a1+="0";
                    }
                    for(int j=i; j<n; j++)
                        a1+=b.charAt(j);
                    
                    prefix.add(1);
                    prefix.add(i+1);
                    count+=2;
                    a=a1;
                }
            }
            
            System.out.print(count+" ");
            for(int i=0; i<prefix.size(); i++)
                System.out.print(prefix.get(i)+" ");
            System.out.println();
            test--;
        }
    }
}