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.*;
import java.util.*;

import static java.lang.Math.*;

public class Main {
    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        int t = nextInt();
        for (int q = 0; q < t; q++) {
            int n = nextInt();
            char[] a = next().toCharArray();
            char[] b = next().toCharArray();
            if (n == 1) {
                if (a[0] != b[0]) pw.println(1 + " " + 1);
                else pw.println(0);
            } else {
                ArrayList<Integer> ans = new ArrayList<>();
                for (int i = 1; i < n; i++) if (a[i] != a[i - 1]) ans.add(i);
                if (a[n - 1] != b[n - 1]) ans.add(n);
                for (int i = n - 1; i >= 1; i--) {
                    if (b[i] != b[i - 1]) ans.add(i);
                }
                pw.print(ans.size() + " ");
                for (int answer : ans) {
                    pw.print(answer + " ");
                }
            }
            pw.println();
        }
        pw.close();
    }

    static long min;
    static long max;
    //    static ArrayList<Pair>[] graph;
    static ArrayList<Integer>[] graph;
    static boolean[] used;

    static void dfs(int v) {
        used[v] = true;
        for (int to : graph[v]) {
            if (!used[to]) dfs(to);
        }
    }

    static long mod = 1_000_000_007;

    static StringTokenizer st = new StringTokenizer("");
    static BufferedReader br;

    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    static long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
}