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 void main(String []args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            int n = sc.nextInt();
            String a = sc.next();
            String b = sc.next();
            ArrayList<Integer> ans = new ArrayList<Integer>();
            char startbit = a.charAt(0);
            int pos1 = 1;
            int pos2 = n - 1;
            int x = 0;
            for(int i = n - 1 ; i >= 0; i--){
                if(b.charAt(i) == startbit)
                    ans.add(1);
                ans.add(i+1);
                x++;
                if(x%2 == 1){
                    startbit = (a.charAt(pos2)=='1')?'0':'1';
                    pos2--;
                }else{
                    startbit = a.charAt(pos1);
                    pos1++;
                }
            }
            n = ans.size();
            System.out.print(n + " ");
            for(int i = 0 ; i < n ; i++)
                System.out.print(ans.get(i)+" ");
            System.out.println();
            
        }
    }
}