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.lang.Math;

//System.out.println();


public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        for (int i = 0; i < n; i++) {
        	int result = 0;
        	StringBuilder strResult = new StringBuilder();
        	int l = sc.nextInt();
        	sc.nextLine();
        	String s1 = sc.nextLine();
        	String s2 = sc.nextLine();
        	int curr = l;
        	boolean flag = true;
        	int ptr = 0;

        	for (int j =0;j<l;j++) {
        		int idx = ptr;
        		if (!flag) {
        			idx = l-ptr-1;
        			ptr++;
        		}

        		char c = s1.charAt(idx);
        		if (c == '1' && !flag) {
        			c = '0';
        		} else if (c == '0' && !flag) {
        			c = '1';
        		}
        		flag = !flag;
        		if (c == s2.charAt(curr-1)) {
        			if (curr-1 == 0) {
        				break;
        			}
        			strResult.append(" 1");
        			// if (s1.charAt(idx) == '1') {
        			// 	s1 = '0' + s1.substring(1);
        			// } else {
        			// 	s1 = '1' + s1.substring(1);
        			// }
        			 strResult.append(" " + curr);
        			 result+=2;
        			// s1 = (new StringBuilder(s1.substring(0, curr)).reverse().toString() + s1.substring(curr)).replace('0', '2').replace('1', '0').replace('2', '1'); 
        			curr--;
        		} else {
        			 strResult.append(" " + curr);
        			// s1 = (new StringBuilder(s1.substring(0, curr)).reverse().toString() + s1.substring(curr)).replace('0', '2').replace('1', '0').replace('2', '1'); 
        			curr--;
        			result++;
        		}
        	}
        System.out.println(result + strResult.toString());	
        }
        
        
    }


}