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.*;
import java.io.*;

public class Task{
// taking inputs
static BufferedReader s1;
static BufferedWriter out;
static String read() throws IOException{String line="";while(line.length()==0){line=s1.readLine();continue;}return line;}
static int int_v (String s1){return Integer.parseInt(s1);}
static long long_v(String s1){return Long.parseLong(s1);}
static int[] int_arr() throws IOException{String[] a=read().split(" ");int[] b=new int[a.length];for(int i=0;i<a.length;i++){b[i]=int_v(a[i]);}return b;}
static long[] long_arr() throws IOException{String[] a=read().split(" ");long[] b=new long[a.length];for(int i=0;i<a.length;i++){b[i]=long_v(a[i]);}return b;}
static void assign(){s1=new BufferedReader(new InputStreamReader(System.in));out=new BufferedWriter(new OutputStreamWriter(System.out));}
static void sort(int[] a){List<Integer> l=new ArrayList<>();for(int z:a){l.add(z);}Collections.sort(l);for(int i=0;i<a.length;i++){a[i]=l.get(i);}}
static long gcd(long a,long b){if(b==0){return a;}return gcd(b,a%b);}
static long Modpow(long a,long p,long m){long res=1;while(p>0){if((p&1)!=0){res=(res*a)%m;}p >>=1;a=(a*a)%m;}return res;}
static long Modmul(long a,long b,long m){return ((a%m)*(b%m))%m;}
static long ModInv(long a,long m){return Modpow(a,m-2,m);}
static int mod=(int)1e9+7;
//https://codeforces.com/blog/entry/53810
//......................................@uthor_Alx..............................................

public static void main(String[] args) throws  IOException{
				assign();
				int t=int_v(read()),cn=1;
				while(t--!=0){
					int n=int_v(read());
					char[] a=read().toCharArray();
					char[] b=read().toCharArray();
					List<Integer> l=new ArrayList<>();
					int j=1,k=0;
					for(int i=n-1;i>0;i--){
						if(a[i]!=b[i]){
							if(a[i]==a[0]){
								l.add(i+1);
								j=0;k=i;
							}
							else{
								l.add(1);l.add(i+1);
								j=1;k=i-1;
							}
								while(j<=k){
									if(a[j]==a[k]){
										a[j]=a[j]=='1'?'0':'1';
										a[k]=a[j];
									}
									j++;k--;
								}
						}
					}
					if(a[0]!=b[0])l.add(1);
					out.write(l.size()+" ");
					for(int x:l){out.write(x+" ");}
					out.write("\n");
				}
			    
			    out.flush();

	}
}