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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author lucasr
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		MyScanner in = new MyScanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		A2PrefixFlipHardVersion solver = new A2PrefixFlipHardVersion();
		int testCount = Integer.parseInt(in.next());
		for (int i = 1; i <= testCount; i++)
			solver.solve(i, in, out);
		out.close();
	}

	static class A2PrefixFlipHardVersion {
		public void solve(int testNumber, MyScanner sc, PrintWriter out) {
			int n = sc.nextInt();
			char[] a = sc.next().toCharArray();
			char[] b = sc.next().toCharArray();
			char[] copy = a.clone();

			int[] ret = new int[2 * n];
			int size = 0, from = 0, to = n - 1;
			boolean reversed = false;

			for (int i = n - 1; i >= 0; i--) {
				if (!reversed) {
					if (a[to] != b[i]) {
						if (a[from] == b[i]) {
							ret[size++] = 1;
							a[from] = other(a[from]);
						}
						ret[size++] = to - from + 1;
						reversed = true;
						from++;
					} else {
						to--;
					}
				} else {
					if (a[from] == b[i]) {
						if (a[to] != b[i]) {
							ret[size++] = 1;
							a[to] = other(a[to]);
						}
						ret[size++] = to - from + 1;
						reversed = false;
						to--;
					} else {
						from++;
					}
				}
			}

			out.print(size);
			for (int i = 0; i < size; i++) {
				out.print(" " + ret[i]);
			}
			out.println();

		}

		static char other(char c) {
			return c == '0' ? '1' : '0';
		}

	}

	static class MyScanner {
		private BufferedReader br;
		private StringTokenizer tokenizer;

		public MyScanner(InputStream is) {
			br = new BufferedReader(new InputStreamReader(is));
		}

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

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

	}
}

