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:
//package CodeForces.RoadMap.D2_B;

import java.util.*;
import java.io.*;
/**
 * @author Syed Ali.
 * @createdAt 17/04/2021, Saturday, 16:47
 */
public class PrefixFlipEasyVersion {
	static class Fs{
			private BufferedReader reader;
			private StringTokenizer stt;

			public Fs() {
				reader = new BufferedReader(new InputStreamReader(System.in));
			}

			public String nextLine() {
				try {
					return reader.readLine();
				} catch (IOException e) {
					return null;
				}
			}

			public String nextString() {
				while (stt == null || !stt.hasMoreTokens()) {
					stt = new StringTokenizer(nextLine());
				}
				return stt.nextToken();
			}

			public int nextInt() {
				return Integer.parseInt(nextString());
			}

			public long nextLong(){ return Long.parseLong(nextString()); }
		}

		static Fs fs = new Fs();

		static void swap(char[] str,int i, int j){
			char temp = (str[i] == '0') ? '1' : '0';
			str[i] = (str[j] == '0') ? '1' : '0';
			str[j] = temp;
		}

		static void reverseAndInvert(char[] str,int start, int end){
			for(int i=start,j = end;i<=j;i++,j--){
				swap(str, i,j);
			}
		}

		static void solve(){
		 	int n = fs.nextInt();
		 	char[] a = fs.nextString().toCharArray();
		 	char[] b = fs.nextString().toCharArray();
		 	List<Integer> prefixLengths = new ArrayList<>();
			int moves = 0;
		 	for(int i=n-1;i>=0;i--){
		 		if(a[i] != b[i]){
		 			if(a[0] != a[i]){
		 				a[0] = a[i];
						prefixLengths.add(1);
		 				++moves;
					}
					++moves;
		 			reverseAndInvert(a,0,i);
					prefixLengths.add(i+1);
				}
			}
			System.out.print(moves + " ");
		 	for(int i=0;i<prefixLengths.size();i++){
				System.out.print(prefixLengths.get(i) + " ");
			}
			System.out.println();
		}

		public static void main(String[] args){
		    int t = fs.nextInt();
		    while(t-->0){
		        solve();
		    }
		}
}