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.util.*;
import java.io.*;

//Solution 1 from editorial
//Credit: The testers Kuroni and SecondThread independently came up with this elegant solution,
//  while solution 2 from editorial was intended.
//We convert a -> 0000000 -> b
//to convert a string to 0000000, we make prefix 2 all same bit, then prefix 3 all same bit, etc.
//to convert 0000000 -> b, we convert b -> 0000000 and reverse the order of operations.

/*
 * Iterate through the bits
 *	When two consecutive bits are not the same, you can change the previous bits
 *	and everything up to the bit will be the same
 *	01001
 *	^
 *	change that bit so that
 *	11001
 *	 ^
 *	this bit is same leftward
 *	000001
 *	    ^
 *	now everything is the same up to this bit
 *	This way we can make every bit 0
 */

public class A1 {
	
	public static void flip(int i, char[] bits) {
		for(int k = 0;k<i;k++) {
			if(bits[k] == '0') bits[k] = '1';
			else bits[k] = '0';
		}
		
		for(int k = 0;k<i/2;k++) {
			char temp = bits[k];
			bits[k] = bits[i-k-1];
			bits[i-k-1] = temp;
		}
	}
	
	public static Scanner stdIO;
	public static Scanner debug;
	public static void main(String[] args) throws IOException{
		stdIO = new Scanner(System.in);
		try {
			debug = new Scanner(new File("debug.txt"));
		}catch(IOException e) {
			
		}
		
		Scanner sc = stdIO;
		
		int t = sc.nextInt();
		for(int i = 0;i<t;i++) {
			int n = sc.nextInt();
			String a = sc.next();
			String b = sc.next();

			Queue<Integer> flipPosA = new LinkedList<Integer>();
			Stack<Integer> flipPosB = new Stack<Integer>();
			a+='0';
			b+='0';
			for(int j = 1;j<=n;j++) {
				if(a.charAt(j) != a.charAt(j-1)) flipPosA.add(j);
				if(b.charAt(j) != b.charAt(j-1)) flipPosB.add(j);
			}
			while(!flipPosB.empty()) flipPosA.add(flipPosB.pop());
			
			if(sc == debug) {
				System.out.println("Demonstrating Flip");
				System.out.println("A:");
				char[] testA = a.toCharArray();
				System.out.println(new String(testA));
				System.out.println("B:\n"+b);
				System.out.println();
				System.out.println(a);
				int s = flipPosA.size();
				for(int j = 0;j<s;j++) {
					int flip = flipPosA.poll();
					flip(flip, testA);
					System.out.println(new String(testA));
					flipPosA.add(flip);
				}
			}
			
			System.out.print(flipPosA.size());
			System.out.print(' ');
			while(!flipPosA.isEmpty()) {
				System.out.print(flipPosA.poll());
				System.out.print(' ');
			}
			System.out.println();
		}
	}
}
