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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class C {

    static BufferedReader br;
    static StringTokenizer st;
    static PrintWriter pw;

    static String nextToken() {
        try {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
        return st.nextToken();
    }

    static int nextInt() {
        return Integer.parseInt(nextToken());
    }

    static long nextLong() {
        return Long.parseLong(nextToken());
    }

    static String nextLine() {
        try {
            return br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }

    static double nextDouble() {
        return Double.parseDouble(nextToken());
    }

    public static void main(String[] args) {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);
        run();
        pw.close();
    }

    private static void run() {
        int t = nextInt();
        for (int i = 0; i < t; i++) {
            int n = nextInt();
            int[] a = new int[n];
            int b[] = new int[n];
            String s = nextToken();
            for (int j = 0; j < n; j++) {
                a[j] = s.charAt(j) - '0';
            }
            s = nextToken();
            for (int j = 0; j < n; j++) {
                b[j] = s.charAt(j) - '0';
            }
            ArrayList<Integer> ans = new ArrayList<>();
            if (a[0] == 1) {
                ans.add(1);
            }
            for (int j = 1; j < n; j++) {
                if (a[j] != 0) {
                    ans.add(j);
                    ans.add(j + 1);

                }
            }
            Arrays.fill(a, 0);
            int k = 0;
            for (int j = n - 1; j >= 0; j--) {
                while (j >= 0 && (a[j] ^ k) == b[j]) {
                    j--;
                }
                if (j >= 0) {
                    ans.add(j + 1);
                    k ^= 1;
                }
            }
            pw.print(ans.size() + " ");
            for (int x :
                    ans) {
                pw.print(x + " ");
            }
            pw.println();
        }
    }
}