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:
// template halnix16


import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {


    static BufferedReader br;

    public static void main(String[] args) throws Exception {
        // write your code here
        br = new BufferedReader(new InputStreamReader(System.in));

        int t = (int) in();

        while (t-- > 0) {
            StringBuilder sb = new StringBuilder();
            int n = (int) in();
            String a1 = sin();
            String a2 = sin();
            int f = 0;
            int s = 0;
            int m = 0;
            StringBuilder ans = new StringBuilder();
            for (int j = n - 1; j >= 0; j--) {

                int ci = -1;
                if(s%2==0){
                    ci = j+f;
                }else{
                    ci= f-j;
                }
                char c = a1.charAt(ci);
                if (s % 2 != 0) {
                    c = (c == '1') ? '0' : '1';

                }
                if (c != a2.charAt(j)) {
                    m+=1;
                    c = a1.charAt(f);
                    if (s % 2 != 0) {
                        c = (c == '1') ? '0' : '1';}

                    if (c == a2.charAt(j)) {
                        m += 1;
                        sb.append(1+" ");

                    }
                        sb.append(j+1+" ");

                    s+=1;f=ci;

                }else{
                }
            }

            System.out.println(m + " " + sb.toString().trim());


        }
    }


    static long[] arr(long n) throws Exception {

        String read[] = sarrin();
        long arr[] = new long[(int) n];
        for (int k = 0; k < n; k++)
            arr[k] = in(read[k]);
        return arr;

    }

    static long[][] arr(long n, long m) throws Exception {


        long arr[][] = new long[(int) n][(int) m];
        for (int k = 0; k < n; k++) {
            String read[] = sarrin();
            for (int j = 0; j < m; j++)
                arr[k][j] = in(read[k]);
        }
        return arr;

    }

    static long in() throws Exception {
        return Long.parseLong(br.readLine());


    }

    static long in(String num) throws Exception {
        return Long.parseLong(num);


    }


    static String sin() throws Exception {

        return br.readLine();
    }

    static String[] sarrin() throws Exception {

        return br.readLine().split(" ");

    }

    static String[] regexSplit() throws Exception {
        return br.readLine().split(" ?(?<!\\G)((?<=[^\\p{Punct}])(?=\\p{Punct})|\\b) ?");
    }


}