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.*;

public class Test {

    private static int r = 1000_000_007;

    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        int t = in.nextInt();
        int n;
        String a, b;
        while (t-- > 0) {
            n = in.nextInt();
            a = in.next();
            b = in.next();
            
            int i = 0;
            
            StringBuffer sb = new StringBuffer();
            int x=0;
            for (int k = 0; k < n; k++) {
                if (k % 2 == 0) {
                    i = k / 2;
                } else {
                    i = n - ((k+1) / 2);
                }
                if (k % 2 == 0) {
                    if (a.charAt(i) != b.charAt(n-1 - k)) {
                        sb.append((n-k)+" ");
                        x++;
                    } else {
                        sb.append(1+" " + (n-k) + " ");
                        x+=2;
                    }
                } else {
                    if (a.charAt(i) == b.charAt(n-1 - k)) {
                        sb.append((n-k)+" ");
                        x++;
                    } else {
                        sb.append(1+ " " + (n-k) + " ");
                        x+=2;
                    }
                }
            }
            System.out.print(x + " ");
            System.out.println(sb);
        }
    }
}

class FastScanner {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

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

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

    int[] readArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

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