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 C {
    public static void main(String[] args) throws IOException{
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int q = sc.nextInt();
        while (q-->0){
            int n = sc.nextInt();
            String a = sc.next();
            StringBuilder res = new StringBuilder("");
            int c = 0;
            String b = sc.next();
            int k = b.length()-1;
            int i = 0;
            boolean t = true;
            for(int j = b.length()-1;j>=0;j--){
                if(t){
                    if(a.charAt(i)==b.charAt(j)){
                        res.append("1 ");
                        c++;
                    }
                    res.append((j+1)+" ");
                    c++;
                    i++;
                }else{
                    if(a.charAt(k)!=b.charAt(j)){
                        res.append("1 ");
                        c++;
                    }
                    res.append((j+1)+" ");
                    c++;
                    k--;
                }
                t = !t;
            }
            pw.println(c + " " +  res);

        }
        pw.flush();
        pw.close();
    }
    static class Pair implements Comparable<Pair>{
        int x;
        int y;
        public Pair(int x,int y){
            this.x= x;
            this.y = y;
        }
        public int compareTo(Pair p){
            if(this.x==p.x)return Long.compare(y,p.y);
            return Long.compare(this.x,p.x);
        }

        @Override
        public String toString() {
            return x+" " + y;
        }
    }
    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

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

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public boolean ready() throws IOException {
            return br.ready();

        }
    }
}