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.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.Collections;
import java.util.ArrayList;

public class _1382C2 {

	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		int T = 1;
		T = fs.nextInt();
		for(int tt = 1; tt <= T; tt++) {
			int n = fs.nextInt();
			char[] a = fs.next().toCharArray();
			char[] b = fs.next().toCharArray();

			ArrayList<Integer> ans = new ArrayList<Integer>();
			boolean flipped = false;
			int l = 0, h = n - 1;

			for(int i = n - 1; i >= 0; i--) {
				char ch = (flipped ? (char)('0' + '1' - a[h]) : a[l]); //a[h] ^ 1
				if(ch == b[i]) {
					ans.add(0);
				}
				ans.add(i);
				if(flipped) {
					h--;
				} else {
					l++;
				}

				flipped ^= true;
			}

			System.out.print(ans.size());
			for(int i: ans) {
				System.out.print(" " + (i + 1));
			}
			System.out.println();
		}
	}

	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		public String next() {
			while (!st.hasMoreElements())
				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()); 
        }
	}
}
