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.util.Scanner;
import java.util.Arrays;

public class PrefixFlip{
	public static void main(String[]args){
		Scanner in = new Scanner(System.in);
		
		int T = in.nextInt();
		int ans[] = new int[2000];
		
		for(int t=0;t<T;t++){
			
			in.nextInt();
			in.nextLine();
			
			char[]chars = in.nextLine().toCharArray();
			char[]rec = in.nextLine().toCharArray();
			
			int n = solve(chars, rec, ans);
			
			System.out.print(n+" ");
			
			for(int i=0;i<n;i++)
				System.out.print(String.valueOf(ans[i])+" ");
			
				
			
			//System.out.println("ans: "+new String(chars));
			
			System.out.println();
		}
	}
	
	static void operate(char in[], int n){
		
		for(int i=0;i<=n;i++)
			in[i] = in[i]=='0'?'1':'0';

		int i=0,j=n;
		
		while(i<j){
			char t = in[i];
			in[i] = in[j];
			in[j] = t;
			i++;
			j--;
		}
		//System.out.println((n+1)+" -> "+new String(in));
	}
	
	static int solve(char[]chars,char[]rec, int []ans){
		int ind=0;
		
		for(int i=chars.length-1;i>=0;i--){
			if(chars[i]!=rec[i]){
				if(chars[0]==rec[i]){
					ans[ind++] = 1;
					chars[0] = chars[0]=='0'?'1':'0';
					//System.out.println((1)+" -> "+new String(chars));
					i++;
				}else{
					operate(chars, i);
					ans[ind++] = i+1;
				}
				
			}
		}
		return ind;
	}
	
}
