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.*;

public class newProb {

	public static void main(String[] args) {
		
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			sc.nextLine();
			StringBuilder a=new StringBuilder();
			a.append(sc.nextLine());
			String b=sc.nextLine();
//			System.out.println(a);
//			System.out.println(b);
			
			List<Integer>al =new ArrayList<>();
			for(int i=n-1;i>=0;i--) {
				if(b.charAt(i)!=a.charAt(i)) {
					if(a.charAt(i)=='0') {
						if(a.charAt(0)=='1') {
							a=func(a,0);
							a=func(a,i);
							al.add(1);
							al.add(i+1);
						}
						else {
							a=func(a,i);
							al.add(i+1);
						}
					}
					else {
						if(a.charAt(0)=='0') {
							a=func(a,0);
							a=func(a,i);
							al.add(1);
							al.add(i+1);
						}
						else {
							a=func(a,i);
							al.add(i+1);
						}
						
					}
					
				}
			}
			
			int l=al.size();
			System.out.print(l+" ");
			for(int e:al) {
				System.out.print(e+" ");
			}
			System.out.println();
		
		}

	}
	static StringBuilder func(StringBuilder s,int r) {
		StringBuilder s2=new StringBuilder();
		s2.append(s.substring(0,r+1));
		for(int i=0;i<s2.length();i++) {
			if(s2.charAt(i)=='1') {
				s2.replace(i, i+1, "0");
			}
			else if(s2.charAt(i)=='0') {
				s2.replace(i, i+1, "1");
			}
		}
		s2.reverse();
		s2.append(s.substring(r+1));
		return s2;
	}

}
