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.*;

public class prefixflipeasy 
{
	public static void main(String[] args)
	{
		Scanner inp = new Scanner(System.in);
		int t = inp.nextInt();
		for(int T=0;T<t;T++)
		{
			int n = inp.nextInt();
			inp.nextLine();
			String a = inp.nextLine();
			String b = inp.nextLine();
			
			int[] aa = new int[n];
			int[] bb = new int[n];
			for(int i=0;i<n;i++)
			{
				if(a.charAt(i)=='0')
					aa[i] = 0;
				else
					aa[i] = 1;
			}
			
			for(int i=0;i<n;i++)
			{
				if(b.charAt(i)=='0')
					bb[i] = 0;
				else
					bb[i] = 1;
			}
			
			int ptr = n-1;
			int ctr = 0;
			int[] ans = new int[2*n];
			
			while(ptr>=0)
			{
				if(aa[ptr]!=bb[ptr])
				{
					if(aa[0]==bb[ptr])
					{
						ans[ctr] = 1;
						ctr++;
						if(aa[0]==0)
							aa[0] = 1;
						else
							aa[0] = 0;
					}
					
					ans[ctr] = ptr + 1;
					ctr++;
					for(int i=0;i<=ptr;i++)
					{
						if(aa[i]==0)
							aa[i] = 1;
						else
							aa[i] = 0;
					}
					for(int i=0;i<(ptr+1)/2;i++)
					{
						int temp = aa[i];
						aa[i] = aa[ptr-i];
						aa[ptr-i] = temp;
					}
				}
				ptr--;
			}
			System.out.print(ctr + " ");
			for(int i=0;i<ctr;i++)
				System.out.print(ans[i] + " ");
			System.out.println();
		}
	}
}