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.BufferedReader;
import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;

public class C658A {
	public static void main(String[] args) throws NumberFormatException, IOException {
		FastReader sc = new FastReader();
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	
		int t = sc.nextInt();
		
		for(int tt=0; tt<t; tt++) {
			int n = sc.nextInt();
			boolean[] a = new boolean[n];
			boolean[] b = new boolean[n];
			
			char[] aa = sc.next().toCharArray();
			char[] bb = sc.next().toCharArray();
			
			for(int i=0; i<n; i++) {
				a[i] = (aa[i] == '1');
				b[i] = (bb[i] == '1');
			}
			
			ArrayList<Integer> ans = new ArrayList<Integer>();
			
			for(int i=0; i<n; i++) {
				int ai=0;
				if(i%2 == 0) {
					ai = i/2;
				} else {
					ai = (n-1) - i/2;
				}
				int bi = (n-1)-i;
				
				if((i%2==0 && a[ai] == b[bi]) || (i%2==1 && a[ai] != b[bi])) {
					ans.add(1);
				}
				ans.add(bi+1);
			}
			
			out.print(ans.size());
			out.print(" ");
			
			for(int an: ans) {
				out.print(an);
				out.print(" ");
			}
			out.println();
		}
		
		out.flush();
	}
	
	static class FastReader{BufferedReader br;StringTokenizer st;public FastReader()
	{br=new BufferedReader(new InputStreamReader(System.in), 32768);}
	String next(){while(st==null||!st.hasMoreElements())
	{try{st=new StringTokenizer(br.readLine());}
	catch(IOException e){e.printStackTrace();}}return st.nextToken();}
	int nextInt(){return Integer.parseInt(next());}
	long nextLong(){return Long.parseLong(next());}
	double nextDouble(){return Double.parseDouble(next());}
	String nextLine(){String str="";try{str=br.readLine();}
	catch(IOException e){e.printStackTrace();}return str;}}
}
