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.ArrayList;
import java.util.List;
import java.util.Scanner;

/**
 * @author 10132
 */
public class Main {
    
    static char[] a, b;
    static int currB;
    static List<Integer> missed = new ArrayList<>(200_000);
    
    static StringBuffer buffer = new StringBuffer(200_000);
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- != 0) {
            int n = sc.nextInt();
            sc.nextLine();
            a = sc.nextLine().toCharArray();
            b = sc.nextLine().toCharArray();
            
            // currB = n;
            // missed.clear();
            // match(0, n - 1, -1, 0);
            // System.out.print(missed.size());
            // for (int i : missed) {
            //     System.out.print(" ");
            //     System.out.print(i);
            // }
            // System.out.println();
            
            System.out.println(String.valueOf(matchIteration(n)) + buffer);
        }
    }
    
    /**
     * Test case 17 TLE
     *
     * @param reduce
     *         a 减小的方向。正向：-1；反向：1
     * @param flip
     *         未翻转：0；翻转：1
     */
    private static void match(int firstA, int lastA, int reduce, int flip) {
        if (--currB < 0) {
            return;
        }
        if ((a[lastA] ^ flip) == b[currB]) {
            match(firstA, lastA + reduce, reduce, flip);
            return;
        }
        
        if ((a[firstA] ^ flip) == b[currB]) {
            missed.add(1);
        }
        missed.add(currB + 1);
        match(lastA, firstA - reduce, -reduce, flip ^ 1);
    }
    
    private static int matchIteration(int n) {
        buffer.setLength(0);
        int missCnt = 0;
        
        int firstA = 0;
        int lastA = n - 1;
        int reduce = -1;
        int flip = 0;
        
        while (--n >= 0) {
            if ((a[lastA] ^ flip) == b[n]) {
                lastA += reduce;
                continue;
            }
            
            if ((a[firstA] ^ flip) == b[n]) {
                missCnt++;
                buffer.append(" ").append(1);
            }
            missCnt++;
            buffer.append(" ").append(n + 1);
            int temp = firstA;
            firstA = lastA;
            lastA = temp - reduce;
            reduce = -reduce;
            flip ^= 1;
        }
        
        return missCnt;
    }
    
}