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.text.*;
import java.util.*;
import java.math.*;

public class CF1382C {
	public static void main(String[] args) throws Exception {
		new CF1382C().run();
	}

	public void run() throws Exception {
		FastScanner f = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		///
		int t = f.nextInt();
		while(t-->0) {
			int n = f.nextInt();
			String as = f.nextLine(), bs = f.nextLine();
			if(as.equals(bs)) { out.println(0); continue; }
			char[] a = as.toCharArray();
			char[] b = bs.toCharArray();
			
			
			List<Integer> aMoves = new ArrayList<>();
			int i = 1;
			while(i < n) {
				if(a[i] != a[i-1]) aMoves.add(i);
				i++;
			}
			if(a[n-1] == '1') aMoves.add(n);
			
			List<Integer> bMoves = new ArrayList<>();
			i = 1;
			while(i < n) {
				if(b[i] != b[i-1]) bMoves.add(i);
				i++;
			}
			if(b[n-1] == '1') bMoves.add(n);
			Collections.reverse(bMoves);
			
			int len = aMoves.size() + bMoves.size();
			out.print(len);
			for(int item : aMoves) out.print(" " + item);
			for(int item : bMoves) out.print(" " + item);
			out.println();
		}
		///
		out.flush();
	}

	///
	static class FastScanner {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public FastScanner() {
			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

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

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

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
	}
}