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.*; 
import java.util.*; 
import java.math.*; 

public class Main { 
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=f.nextInt();
//		int t=1;
		while(t-->0){
			int n=f.nextInt();
			char[] a=f.next().toCharArray();
			char[] b=f.next().toCharArray();
			
			int ind=n-1;
			int start=0;
			int end=n-1;
			boolean flag=false;
			ArrayList<Integer> ans=new ArrayList<>();
			while(ind>-1) {
				if(a[end]==b[ind] && !flag) {
					ind--;
					end--;
					
				}
				else if(a[start]!=b[ind] && flag) {
					ind--;
					start++;
				}
				else {
					if(flag==false) {
						if(a[start]==a[end]) {
							ans.add(ind+1);
							flag=true;
							ind--;
							start++;
						}
						else {
							ans.add(1);
							ans.add(ind+1);
							flag=true;
							ind--;
							start++;
						}
					}
					else {
						if(a[end]==a[start]) {
							ans.add(ind+1);
							flag=false;
							ind--;
							end--;
						}
						else {
							ans.add(1);
							ans.add(ind+1);
							flag=false;
							ind--;
							end--;
						}
					}
				}
//				System.out.println(ind+" "+start+" ");
				
			}
			System.out.print(ans.size()+" ");
			for(int i=0;i<ans.size();i++) {
				System.out.print(ans.get(i)+" ");
			}
			System.out.println();
		}
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
	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());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 