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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.StringTokenizer;

public class Main {

    static final long M = 1000000007;

    public static void main(String args[]) {

        FastReader io = new FastReader();

        int t = io.nextInt();

        while(t > 0){
            int n = io.nextInt();
            String a = io.nextLine();
            String b = io.nextLine();

            new Solver().solve(n,a,b);

            t--;
        }
    }
}

class Solver{
    static final long MAX = 10000000000L;
    public void solve(int n,String a,String b){

        char A[] = a.toCharArray();
        char B[] = b.toCharArray();

        ArrayList<Integer> ans = new ArrayList<>();
        int i = 0,j = 0;
        while(j < n){

            if(j%2 == 0){
                char first = (char)((((A[i] - 48) + j)%2) + 48);
                if(B[n - j - 1] == first){
                    ans.add(1);
                }
                i++;
            }
            else{
                char first = (char)((((A[n - i] - 48) + j)%2) + 48);
                if(B[n - j - 1] == first) {
                    ans.add(1);
                }
            }

            ans.add(n - j);

            j++;
        }

        System.out.print(ans.size() + " ");

        for (int k = 0; k < ans.size();k ++){
            System.out.print(ans.get(k) + " ");
        }

        System.out.println();

    }
}

class Util{
    static  final  long M = 1000000007;
    static long modexp(long n,long m){
        if(m==0)
            return 1;
        else if(m==1)
            return n;
        else{
            long p = modexp(n,m/2);

            if(m%2==1)
                return (((p*p)%M)*n)%M;
            else
                return (p*p)%M;
        }
    }

    static  long exp(long n,long m){
        if(m == 0)
            return 1;
        if(m == 1)
            return n;

        long p = exp(n,m/2);

        if(m%2 == 1)
            return p*p*n;
        else
            return p*p;
    }

    static long gcd(long a,long b){
        if(a==0)
            return b;

        return gcd(b%a,a);
    }

    static long inv(long n){
        return modexp(n,M-2);
    }

    static long lcm(long a,long b){
        return a*b/gcd(a,b);
    }
    static long factorial(long fact[],long n){
        fact[0]=1;
        fact[1]=1;

        long prod = 1;

        for(int i = 2;i <= n;i++){
            prod = (prod*i)%M;

            fact[i] = prod;
        }

        return prod;
    }

    static boolean isPrime(long n){
        for(int i = 2;i*i <= n;i++){
            if(n%i == 0)
                return false;
        }
        return true;
    }
}

class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader() {
        br = new BufferedReader(new
                InputStreamReader(System.in));
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            }
            catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        }
        catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}