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.util.Scanner;

public class C1 {

	public static void main(String[] args) {

		Scanner nik = new Scanner(System.in);
		int t = nik.nextInt();
		StringBuilder st = new StringBuilder();
		while (t-- > 0) {

			int n = nik.nextInt();
			char[] c1 = nik.next().toCharArray();
			char[] c2 = nik.next().toCharArray();
			int c = 0;

			StringBuilder st1 = new StringBuilder();

			for (int i = n - 1; i >= 0; i--) {

				if (c1[i] != c2[i]) {

					if(c1[0]==c2[i]) {
						c++;
						c1[0] = c1[0]=='0'?'1':'0';
						st1.append(1+" ");
					}
					int temp = i+1;
					st1.append(temp+" ");
					c++;
					swap(c1,0,i);
				}

			}
			
//			for(char val:c1) {
//				System.out.print(val+" ");
//			}
//			System.out.println();
//
//			for(char val:c2) {
//				System.out.print(val+" ");
//			}
//			System.out.println();
			
			st.append(c + " " + st1 + "\n");

		}
		System.out.println(st);

	}

	private static void swap(char[] c1, int i, int i2) {

		while(i<i2) {
			
			char temp = c1[i];
			c1[i] = c1[i2];
			c1[i2] = temp;
			c1[i] = c1[i]=='0'?'1':'0';
			c1[i2] = c1[i2]=='0'?'1':'0';
			i++;
			i2--;
		}
		if(i==i2) {
		 c1[i] = c1[i]=='0'?'1':'0';
		}
		
	}

}