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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;



import java.util.*;





public class MainB {
	public static String solve(String s) {
		//System.out.println(s);
		char c[]=s.toCharArray();
		int n=s.length();
		String ans="";
		for(int i=0;i<n;i++) {
			if(c[i]=='0') c[i]='1';
			else c[i]='0';
			
			
		}
		ans=String.valueOf(c);
		//System.out.println("ans "+ans);
		StringBuilder sBuilder=new StringBuilder(ans);
		sBuilder.reverse();
		
		
		return String.valueOf(sBuilder);
		
		
		
	}
	
	public static void main(String[] args) {
		FastScanner scanner=new FastScanner();
		int t=scanner.nextInt();
		while(t-->0) {
			int n=scanner.nextInt();
			String a=scanner.next(),b=scanner.next();
			ArrayList<Integer> al=new ArrayList<>();
			char c1[]=a.toCharArray();
			char c2[]=b.toCharArray();
			
			for(int i=0;i<n;i++) {
				if(c1[i]=='1') {
				if(i>0) al.add(i);
				al.add(i+1);
				}
			}
			for(int i=n-1;i>=0;i--) {
				if(c2[i]=='1') {
					al.add(i+1);
				if(i>0) al.add(i);
				
				}
			}
			if(al.size()>2*n) {
				al.clear();
				for(int i=0;i<n;i++) {
					if(c1[i]=='0') {
					if(i>0) al.add(i);
					al.add(i+1);
					}
				}
				for(int i=n-1;i>=0;i--) {
					if(c2[i]=='0') {
						al.add(i+1);
					if(i>0) al.add(i);
					
					}
				}
				
			}
			
			
			
			System.out.print(al.size());
			
			for(int i:al)
				System.out.print(" "+i);
			
			System.out.println();
			
				
			
		}
					
		}

	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
}

}
