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.
/*
     D    A    R    K    L      _                 _         O    R    D    D    A
         K                    / |                 | \        L    O    R         D
        A         R         _ / |                 | \ _       K    L         O    R
  D         A    R    K     _ / |         _       | \ _        L    O    R    D    D
      R    K    L       _ / _   |       _ | \\    |   _ \       O    R    D    D
     R         K        _ / _   |       _ | \ \   |   _ \        L    O    R
    D    A    R     _ / _ / |   /     /         \ \   | \ _ \ _        K    L    O
   D    D           _ / _ / |   /     /          \ \   | \ _ \ _             A
       K        _ / _ /     | /     /     _       \ \ |     \ _ \ _      L    O    R
 D    A        _ / _ /      | /     /     _       \ \ |      \ _ \ _      R         K
            _ / _ /         | |     |   |   \ (O) / | |         \ _ \ _         O
    D      _ / _ /          | |     |   |    \   /  | |          \ _ \ _         D
   A      / _ /     |   |   | \     |   \ _   \|/   / |   |   |     \ _ \
  K      / _ /      |   |   | \     |   \ _    V    / |   |   |      \ _ \         L
        / /         |   |   |   \ _ /       \ _ _ /   |   |   |         \ \
       / /          |   |   |                         |   | | |          \ \
      / /       |   | | \     \       ROWECHEN      /     / | |   |       \ \
     / /    _ _ |   | | \     \        ZHONG        /     / | |   | _ _    \ \
    / / _ /   \ | | | |   \ / \ \                 / / \ /   | | | | /   \ _ \ \
   / / _ /    \ | | | |       \ \                 / /       | | | | /    \ _ \ \
  / / /         \ \ \ \ /       /                 \       \ / / / /         \ \ \
 \ / /          \ \ \ \ /       /                 \       \ / / / /          \ \ /
  \|/             \|/         |         |         |           \|/             \|/
  L         O     \|/         |         |   |     |           \|/        R    D
 A    R    K    L             / _ |     |   | _     \      O    R    D    D    A    R
     L    O         R         / _ |     |   | _     \       D    D         A    R
    L    O    R    D        / / / _     | _ | | \ _   \      D    A    R         K
   O    R    D              / / / _ |   | _ | | \ _   \       D    A    R    K    L
  R    D    D    A    R     | / / | | / | | \ / | | \ |        K    L    O    R    D
 A         R         K      | / / | | / | | \ / | | \ |              L    O         R
     D    A         R         /   \ / | | | / | | / \ /     K    L         O    R
         D    D         A     /   \ / | | | / | | / \/            R    K    L    O
             R         D    D     / | / \ |   \ /        A    R    K    L    O    R
       D    A    R    K    L      / | / \ |    \/              O    R    D    D
 R    K    L    O    R    D       \ /     |      D    A    R         K    L    O    R
          D    A    R             \/      |            K    L    O    R    D    D

*/
//TEMPLATE V2

import java.io.*;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class A1381_2 {
//Solution goes below: ------------------------------------
public static void solution() throws IOException{
    int t= nextInt();
    for(int test=0;test<t;test++){
        int n = nextInt();
        char[] ac = nextLine().toCharArray();
        char[] bc = nextLine().toCharArray();
        boolean[] a = new boolean[n],b = new boolean[n];
        for(int i=0;i<n;i++){
            a[i] = ac[i]=='1';
            b[i] = bc[i]=='1';
        }
        Arr<Integer> res = new Arr<>();
        int l = 0;
        int dir = -1;
        // We're going forward, which means we're backwards.
        for(int i = n-1;i>=0;i--)
            if (dir == 1) if (a[l] == b[i]) { //Reversed.
                if (a[l] != a[l + i]) res.add(0);
                res.add(i);
                dir = -1;
            } else l += 1;
            else if (a[l + i] != b[i]) {
                if (a[l + i] != a[l]) res.add(0);
                res.add(i);
                l += 1;
                dir = 1;
            }
        print(res.size());
        for(int i:res){
            print(" ");print(i+1);
        }
        println();
    }
}

//Solution goes above: ------------------------------------
public static final String IN_FILE = "";
public static final String OUT_FILE = "";
//-------------------- ------------------------------------
//IO
public static BufferedReader br;
public static StringTokenizer st;
public static BufferedWriter bw;
public static void main(String[] args) throws IOException{
    if(IN_FILE=="") br = new BufferedReader(new InputStreamReader(System.in));
    else try {
        br = new BufferedReader(new FileReader(IN_FILE));
    } catch (FileNotFoundException e) {
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    if (OUT_FILE=="") bw = new BufferedWriter(new OutputStreamWriter(System.out));
    else try {
        bw = new BufferedWriter(new FileWriter(OUT_FILE));
    } catch (FileNotFoundException e) {
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
    }
    solution();
    bw.close();//Flushes too.
}

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

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

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

public static void println() throws IOException{
    bw.newLine();
}

public static void print(Object s) throws IOException{
    bw.write(s.toString());
}
//Other
public static class Arr<T> extends ArrayList<T> {} //I hate typing ArrayList
}