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.util.*;
import java.lang.*;
import java.io.*;

	public class TestClass {
		
		BufferedReader br;
		StringTokenizer st;
		
		public TestClass() { // constructor
			
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() {
			
			while (st == null || !st.hasMoreElements()) {
				try{
					st = new StringTokenizer(br.readLine());
				}
				
				catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt() {
			
			return Integer.parseInt(next());
		}
		
		long nextLong () {
			
			return Long.parseLong(next());
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			
			String str = "";
			try{
				
				str = br.readLine();
			}
			catch (IOException e) {
				
				e.printStackTrace();
			}
			return str;
		}
		
		static void sort(int[] arr) {
			
	        int n = arr.length;
	        
	        Random rnd = new Random();
	        
	        for(int i = 0; i < n; ++i) {
	        	
	            int tmp = arr[i];
	            int randomPos = i + rnd.nextInt(n - i);
	            arr[i] = arr[randomPos];
	            arr[randomPos] = tmp;
	        }
	        
	        Arrays.sort(arr);
	   }
		
		static boolean isPrime(int n) {
			for(int i = 2; i * i <= n; i++) {
				if(n % i == 0) {
					return false;
				}
			}
			return true;
		}
		
	public static void main(String[] args) throws Exception{
		
		TestClass in = new TestClass();
		PrintWriter out = new PrintWriter(System.out);
		
		int test = in.nextInt();
		
		while(test-- > 0) {
			
			int n = in.nextInt();
			
			String a = in.next();
			String b = in.next();
			
			char ch[] = a.toCharArray();
			int ans = 0;
			
			StringBuilder sb = new StringBuilder();
			
			for(int i = n - 1; i >= 0; i--) {
				if(ch[i] != b.charAt(i)) {
					if(ch[0] == b.charAt(i)) {
						ans++;
						sb.append(1 + " ");
						swap(ch, 0, 0);
					}
					 
					ans++;
					sb.append((i + 1) + " ");
					swap(ch, 0, i);
				}
			}
			
			out.println(ans + " " + sb.toString());
		}
		
		out.flush();
		out.close();
	}
	static void swap(char ch[], int i, int j) {
		
		while(i < j) {
			if(ch[i] == '0') {
				ch[i] = '1';
			}
			else {
				ch[i] = '0';
			}
			if(ch[j] == '0') {
				 ch[j] = '1';
			}
			else {
				ch[j] = '0';
			}
			
			char temp = ch[i];
			ch[i] = ch[j];
			ch[j] = temp;
			i++;
			j--;
		}
		
		if(i == j) {
			if(ch[i] == '0') {
				ch[i] = '1';
			}
			else {
				ch[i] = '0';
			}
		}
		
	}
}