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.io.*;
import java.lang.Math;


public class Solution{

	public static void main(String[] args) throws Exception{
		PrintWriter out = new PrintWriter(System.out);

		FastScanner fs = new FastScanner();
		int t = fs.nextInt();

		while(t-->0){

			int n = fs.nextInt();
			char[] a = fs.next().toCharArray();
			char[] b = fs.next().toCharArray();

			LinkedList<Integer> list1 = new LinkedList<Integer>();
			LinkedList<Integer> list2 = new LinkedList<Integer>();

			int n1 = 0;

			//NUM OF WAYS TO CONVERT A TO 000000
			int c = a[0];
			for(int i=0;i<n;i++){
				if(a[i]!=c){
					n1++;
					list1.add(i);
					c = a[i];
				}
			}

			if(a[n-1]!='0'){
				n1++;
				list1.add(n);
			}


			
			int n2 = 0;

			int ch = b[0];

			//NUM OF WAYS TO CONVERT B TO 0000 AND THEN WE WILL REVERSE IT
			for(int i=0;i<n;i++){
				if(b[i]!=ch){
					n2++;
					list2.addFirst(i);
					ch = b[i];
				}
			}

			if(b[n-1]!='0'){
				n2++;
				list2.addFirst(n);
			}


			out.print((n1+n2)+" ");
			for(int d: list1)
				out.print(d+" ");

			for(int d: list2)
				out.print(d+" ");

			out.println("");

			
		}
		

		
		out.close();

		
	}

	




	


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

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

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

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

}