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.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.time.LocalTime;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;


public class C {


    public static void main(String[] args) {
    
    	FastScanner scan = new FastScanner();
    	
    	int t = scan.nextInt();
    	for(int tt=0; tt<t; tt++) {
    			int n = scan.nextInt();
    			char [] ch = scan.next().toCharArray(), b = scan.next().toCharArray();
    			ArrayList<Integer> arr = new ArrayList<>();
    			
    			for(int i=n-1; i>=0; i--) {
    				if(ch[i] != b[i]) {
    					if(ch[0] != ch[i]) {
    						if(ch[0] == '1') ch[i] = '0';
    						else ch[0] = '1';
    						arr.add(1);
    					}
    					ch = swap(i, ch);
    					arr.add(i+1);
    				}
    			}
    			
    			if(arr.size() == 0) System.out.println("0");
    			else {
    				System.out.print(arr.size() + " " );
    				for(int i=0; i<arr.size(); i++) 
    				System.out.print(arr.get(i) + " ");
    				System.out.println();
    			}
    	}	
   }
    
    static char[] swap(int j, char [] ch) {
    	int n = ch.length;
    	char [] ans = new char[n];
    	int place = j;
    	for(int i=0; i<=j; i++) {
    		char a = ch[i];
    		if(a == '1') a = '0';
    		else a = '1';
    		ans[place--] = a;
    	}
    	for(int i=j+1; i<n; i++) {
    		ans[i] = ch[i];
    	}
    	return ans;
    }
  
    	
    public static void sort(int [] a) {
    	ArrayList<Integer> b = new ArrayList<>();
    	for(int i: a) b.add(i);
    	Collections.sort(b);
    	for(int i=0; i<a.length; i++) a[i]= b.get(i);
    }
    static class FastScanner{
    	
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer("");
    	String next() {
    		while(!st.hasMoreTokens())
    			try {
    				st = new StringTokenizer(br.readLine());
    			} catch (IOException e){
    				e.printStackTrace();
    			}
    		return st.nextToken();
    	}
    	
    	int nextInt() {
    		return Integer.parseInt(next());
    	}
    	
    	int [] readArray(int n) {
    		int [] a = new int[n];
    		for(int i=0; i<n ; i++) a[i] = nextInt();
    		return a;
    	}
    	
    	long nextLong() {
    		return Long.parseLong(next());
    	}
    	
    	
    }
    
   
}