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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class B {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			char[] a = sc.next().toCharArray();
			char[] b = sc.next().toCharArray();
			solve(a, b, n, out);
		}
		out.close();
	}

	private static void solve(char[] a, char[] b, int n, PrintWriter out) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i = n - 1; i >= 0; i--) {
			if (a[i] != b[i]) {
				if (b[i] == a[0]) {
					l.add(0);
					a[0] = a[0] == '1' ? '0' : '1';
				}
				l.add(i);
				change(a, i);
			}
		}
		out.print(l.size() + " ");
		for (int i : l) {
			i++;
			out.print(i + " ");
		}
		out.println();
	}

	private static void change(char[] a, int i) {
		int s = 0;
		int e = i;
		while (s <= e) {
			a[s] = a[s] == '1' ? '0' : '1';
			if (s != e) {
				a[e] = a[e] == '1' ? '0' : '1';
			}
			swap(a, s, e);
			s++;
			e--;
		}
	}

	private static void swap(char[] a, int s, int e) {
		char temp = a[s];
		a[s] = a[e];
		a[e] = temp;
	}

	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

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

		int[] readArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

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

}
