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 C {

	static long m = (long) (1e9 + 7);

	public static void main(String[] args) throws IOException {
		Scanner scn = new Scanner(System.in);
		StringBuilder sb = new StringBuilder();
		int T = scn.nextInt(), tcs = 0;
		B: while (tcs++ < T) {
			int n = scn.nextInt();
			char[] a = scn.next().toCharArray();
			char[] b = scn.next().toCharArray();
			int c = 0;

			StringBuilder st1 = new StringBuilder();

			for (int i = n - 1; i >= 0; i--) {
				if (a[i] != b[i]) {
					if (a[0] == b[i]) {
						c++;
						a[0] = a[0] == '0' ? '1' : '0';
						st1.append(1 + " ");
					}
					int temp = i + 1;
					st1.append(temp + " ");
					c++;
					swap(a, 0, i);
				}
			}
			sb.append(c + " " + st1);
			sb.append("\n");
		}
		System.out.print(sb);
	}

	private static void swap(char[] aa, int i, int i2) {
		while (i < i2) {
			char temp = aa[i];
			aa[i] = aa[i2];
			aa[i2] = temp;
			aa[i] = aa[i] == '0' ? '1' : '0';
			aa[i2] = aa[i2] == '0' ? '1' : '0';
			i++;
			i2--;
		}
		if (i == i2) {
			aa[i] = aa[i] == '0' ? '1' : '0';
		}
	}

	static class Scanner {

		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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