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 file {

    private static final Scanner scn = new Scanner(System.in);

    private static void doOp(char[] a, int n) {
        
        for (int i = 0; i < n/2; i++) {
            char tmp = a[i];
            a[i] = a[n-i-1];
            a[n-i-1] = tmp;
        }
        for (int i = 0; i < n; i++) {
            a[i] = a[i] == '1' ? '0' : '1';
        }
    }

    public static void foo(int k){
        int n = scn.nextInt();
        String a = scn.next();
        String b = scn.next();
        char[] aaa = a.toCharArray();
        char[] bbb = b.toCharArray();
        ArrayList<Integer> lis = new ArrayList<>();
        for (int i = b.length()-1; i >= 0; i--) {
            char chk = b.charAt(i);
            if (chk != aaa[i]) {
                if (chk == aaa[0]) {
                    doOp(aaa, 1);
                    lis.add(1);
                } 
                doOp(aaa, i+1);
                lis.add(i+1);
            }
        }
        System.out.print(lis.size() + " ");
        for (int aa : lis) {
            System.out.printf("%d ", aa);
        }
        System.out.println();
    }


    public static void main(String[] args) {

        int t = scn.nextInt();
        for (int i = 0; i < t; i++) {
            foo(i);
        }
    }
}