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.*;
import java.util.*;
 
public class Main {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter pw = new PrintWriter(System.out);
		
		int Q = sc.ni();
		for (int q = 0; q < Q; q++) {
			int N = sc.ni();
			String S = sc.next();
			String T = sc.next();
			int[] s = new int[N];
			for (int i = 0; i < N; i++) {
				s[i] = S.charAt(i)-'0';
			}
			int[] t = new int[N];
			for (int i = 0; i < N; i++) {
				t[i] = T.charAt(i)-'0';
			}
			
			int L = 0;
			int R = N-1;
			boolean rev = false;
			boolean flip = false;
			ArrayList<Integer> moves = new ArrayList<Integer>();
			for (int i = N-1; i >= 0; i--) {
				int end = s[R];
				if (rev)end = s[L];
				if (flip)end = 1-end;
				if (end == t[i]) {
					if (rev)
						L++;
					else
						R--;
				} else {
					int beg = s[L];
					if (rev)beg = s[R];
					if (flip)beg = 1-beg;
					if (beg == t[i]) {
						moves.add(0);
					}
					//reverse and flip
					moves.add(R-L);
					rev = !rev;
					flip = !flip;
					if (rev) {
						L++;
					} else {
						R--;
					}
				}
			}
			pw.print(moves.size() + " ");
			for (int m: moves)
				pw.print((m+1) + " ");
			pw.println();
		}
		
		pw.close();
	}
    
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next() { 
            while (st == null || !st.hasMoreElements()) { 
                try { 
                    st = new StringTokenizer(br.readLine());
                } catch (IOException  e) { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        }
        
        int ni() { 
            return Integer.parseInt(next()); 
        }
  
        long nl() { 
            return Long.parseLong(next()); 
        } 
  
        double nd() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() {
            String str = ""; 
            try { 
                str = br.readLine(); 
            } catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str;
        }
    }
}