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.

import java.io.*;

public class Prefix_Flip {
    public static void main(String[] args) throws java.lang.Exception{
        BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(input.readLine());
        for(int i=0;i<t;i++){
            int n = Integer.parseInt(input.readLine());
            char[] c1 = input.readLine().toCharArray();
            char[] c2 = input.readLine().toCharArray();
            String ans = "";
            int count = 0;
            for(int j=n-1;j>=0;j--){
                if(c2[j]==c1[j]){
                    continue;
                }else if(c2[j]!=c1[0]){
                    op(c1, j);
                    count++;
                    ans += Integer.toString(j+1)+" ";
                    continue;
                }else {
                    op(c1,0);
                    ans += Integer.toString(1)+" ";
                    count++;
                    op(c1,j);
                    ans += Integer.toString(j+1)+" ";
                    count++;
                }
            }
            /*for(int j=0;j<c1.length;j++){
                System.out.print(c1[j]);
            }
            System.out.println();*/
            System.out.println(count+" "+ans);
        }
    }

    public static void op(char[] c, int a){
        for(int i=0;i<=a;i++){
            if(c[i]=='0'){
                c[i] = '1';
            }else {
                c[i] = '0';
            }
        }
        int x = 0;
        int y = a;
        while(x<y){
            char temp = c[x];
            c[x] = c[y];
            c[y] = temp;
            x++;
            y--;
        }
    }
}
