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 scanner = new Scanner(System.in);
            int t = scanner.nextInt();
            while (t-- > 0) {
                int n = scanner.nextInt();
                int[] a = new int[n];
                ArrayList<Integer> ans = new ArrayList<>();
                String a_str = scanner.next();
                for (int i = 0; i < n; ++i) {
                    a[i] = a_str.charAt(i) - '0';
                }
                int[] b = new int[n];
                String b_str = scanner.next();
                for (int i = 0; i < n; ++i) {
                    b[i] = b_str.charAt(i) - '0';
                }
                for (int i = n - 1; i >= 0; --i) {
                    if (a[i] != b[i]) {
                        if (a[0] != a[i]) {
                            changeUp(a, 0);
                            ans.add(1);
                        }
                        changeUp(a, i);
                        ans.add(i + 1);
                    }
                }
                System.out.print(ans.size() + " ");
                for (int x : ans)
                    System.out.print(x + " ");
                System.out.println();
            }
        }

        static void changeUp(int[] arr, int pre) {
            for (int i = 0, j = pre; i < j; ++i, --j) {
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
            for (int i = 0; i <= pre; ++i) {
                if (arr[i] == 0)
                    arr[i] = 1;
                else
                    arr[i] = 0;
            }
        }
    }