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.*;
import java.util.*;

public class Main {
		
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int tc = in.nextInt();
		for(int pt=0;pt<tc;pt++) {
			int n = in.nextInt();
			String a = in.next();
			String b = in.next();
			ArrayList<Integer> ans = new ArrayList<Integer>();
			int first = 0;
			int last = n-1;
			for(int i=0;i<n;i++) {
				int index = i;
				if(i%2==0) {
					index = first;
					first++;
					if(a.charAt(index) == b.charAt(n-1-i)) {
						ans.add(1);
						ans.add(n-i);
					}
					else {
						ans.add(n-i);
					}
				}
				else {
					index = last;
					last--;
					if(a.charAt(index) != b.charAt(n-1-i)) {
						ans.add(1);
						ans.add(n-i);
					}
					else {
						ans.add(n-i);
					}
				}
				
			}
			System.out.print(ans.size()+" ");
			for(int k:ans) {
				System.out.print(k+" ");
			}
			System.out.println();
//			String s = flip(a,ans);
//			System.out.println("String: "+s);
//			System.out.println(s.equals(b));
		}
	}
//
//	private static String flip(String a, ArrayList<Integer> ans) {
//		// TODO Auto-generated method stub
//		String p = a;
//		int n = a.length();
//		for(int k:ans) {
//			String l = "";
//			String pt = "";
//			int i=0;
//			while(i<k) {
//				if(p.charAt(i)=='0') {
//					l+="1";
//				}
//				else {
//					l+="0";
//				}
//				i++;
//			}
//			pt = reverse(l);
//			while(i<n) {
//				pt+=p.charAt(i);
//				i++;
//			}
//			p = pt;
//		}
//		return p;
//	}
//
//	private static String reverse(String l) {
//		// TODO Auto-generated method stub
//		String s = "";
//		for(int i=l.length()-1;i>=0;i--) {
//			s+=l.charAt(i);
//		}
//		return s;
//	}
}