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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.StringTokenizer;


public class C {
	
	
	
	public static void main(String[] args) throws IOException {
		FastScanner sc=new FastScanner();
		int t = sc.nextInt();
		while(t-- != 0) {
			int n = sc.nextInt();
			char[] str = sc.next().toCharArray();	
			char[] str1 = sc.next().toCharArray();	

			int arr[] = new int[n];
			int arr1[] = new int[n];
			for(int i=0; i<n; i++)
				arr[i] = str[i] == '1' ? 1 : 0;
			for(int i=0; i<n; i++)
				arr1[i] = str1[i] == '1' ? 1 : 0;
			ArrayList<Integer> lis = new ArrayList<Integer>();
			
			int i=0;
			int j = n-1;
			int k = n-1;
			
			while(j>=0 && k>=0) {
				if(arr1[j] == arr[k]) {
					j--;
					k--;
					continue;
				}
				
				if(arr[i] == arr1[j]) {
					lis.add(1);
					arr[i] = arr[i] ^ 1;
					lis.add(j+1);
					swap(arr,j);
				}
				else {
					lis.add(j+1);
					swap(arr,j);
				}
				
			}
			
			System.out.print(lis.size()+" ");
			for(int e : lis) {
				System.out.print(e+" ");
			}
			System.out.println();
			
			
			
			
		}
		
		
	}
	
	
	
	
 
	private static void swap(int[] arr, int j) {
		
		int index = j;
		int temp[] = new int[arr.length];
		for(int i=0; i<arr.length; i++)
			temp[i] = arr[i];
		
		for(int i=0; i<=j; i++) {
			arr[i] = temp[index] ^ 1;
			index--;
		}
		
	}





	static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.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());
		}
	}
}