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.PrintWriter;
import java.util.*;

public class Main {
    static boolean inversed;
    static int l = 0;
    static int n;
    static int id1, id2;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();
        while (t-- > 0) {
            id1 = 1;
            id2 = 1;
            n = in.nextInt();
            char[] a = in.next().toCharArray();
            char[] b = in.next().toCharArray();
            l = 0;
            int r = n - 1;
            List<Integer> ops = new ArrayList<>();
            inversed = false;
            while (r >= 0) {
                char al = a[l];
                if (inversed) {
                    al = inverseChar(al);
                }
                if (al == b[r]) {
                    if (r == 0) {
                        break;
                    }
                    a[l] = inverseChar(al);
                    if (inversed) {
                        a[l] = inverseChar(a[l]);
                    }
                    ops.add(1);
                }
                execute(a, r);
                ops.add(r + 1);
                r--;
            }
            out.print(ops.size());
            for (int x : ops) {
                out.print(" " + x);
            }
            out.println();
        }
        out.close();
    }

    static char inverseChar(char c) {
        return (char) (('1' - c) + '0');
    }

    static void execute(char[] a, int p) {
        if (!inversed) {
            l = n - id1++;
        } else {
            l = id2++;
        }
        inversed = !inversed;
    }
}
