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.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;


public class Solution {
	public static class FastReader {
			BufferedReader br;
			StringTokenizer root;
			
	 
			public FastReader() {
				br = new BufferedReader(new InputStreamReader(System.in));
			}
	 
			String next() {
				while (root == null || !root.hasMoreTokens()) {
					try {
						root = new StringTokenizer(br.readLine());
					} catch (Exception addd) {
						addd.printStackTrace();
					}
				}
				return root.nextToken();
			}
	 
			int nextInt() {
				return Integer.parseInt(next());
			}
	 
			double nextDouble() {
				return Double.parseDouble(next());
			}
	 
			long nextLong() {
				return Long.parseLong(next());
			}
	 
			String nextLine() {
				String str = "";
				try {
					str = br.readLine();
				} catch (Exception addd) {
					addd.printStackTrace();
				}
				return str;
			}
		}
	 
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	public static FastReader sc = new FastReader();
 
	static int mod = (int) (1e9+7),MAX=(int)(1e9);

	static StringBuilder str;
	public static void main(String[] args) throws IOException{

    	int t = sc.nextInt();
		while(t-->0) {
			 int n = sc.nextInt();
			 char[] a = sc.next().toCharArray();
			 char[] b = sc.next().toCharArray();
			 List<Integer> list = new ArrayList<>();
			 List<Integer> temp = new ArrayList<>();
			 char c = a[0];
			 for(int i=1;i<a.length;++i) {
				 if(a[i] != c) {
					 list.add(i);
					 c = a[i];
				 }
			 }
			 
			 c = b[0];
			 for(int i=1;i<b.length;++i) {
				 if(b[i] != c) {
					 temp.add(i);
					 c = b[i];
				 }
			 }
			 if(a[n-1] == '1') list.add(n);
			 if(b[n-1] == '1') temp.add(n);
			 int op = list.size() + temp.size();
			 out.print(op+" ");
			 for(int i=0;i<list.size();++i) out.print(list.get(i)+" ");
			 for(int i=temp.size()-1;i>=0;--i) out.print(temp.get(i)+" ");
			 out.println();
			  
		}
		out.close();
	}
	

}