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 Solution {
	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int t = Integer.parseInt(st.nextToken());
		PrintWriter out = new PrintWriter(System.out);
		for(int qq=0;qq<t;qq++){
		    
		    st = new StringTokenizer(br.readLine());
		    int n = Integer.parseInt(st.nextToken());
		    st = new StringTokenizer(br.readLine());
		    char[] a = st.nextToken().toCharArray();
		    st = new StringTokenizer(br.readLine());
		    char[] b = st.nextToken().toCharArray();
		    //if(qq!=1) continue;
		    int beg = n-1;
		    int end = 0;
		    int ind = n-1;
		    int ste = 0;//num+ste % 2
		    LinkedList<Integer> al = new LinkedList<Integer>();
		    while(ind>=0){
		        
		        if(((int)(a[beg]-'0')+ste)%2==(int)(b[ind]-'0')){
		            ind--;
		            if(ste==0) beg--;
		            else beg++;
		        }else{
		            //System.out.println(ind+" "+((int)(a[beg]-'0')+ste)%2+" "+(int)(b[ind]-'0'));
		            //System.out.println(((int)(a[end]-'0')+ste)%2+" "+((int)(a[beg]-'0')+ste)%2);
		            if(((int)(a[end]-'0')+ste)%2!=((int)(a[beg]-'0')+ste)%2){
		                al.add(1);
		            }
		            int size = beg - end;
		            if(size<0) size *= -1;
		            size++;
		            al.add(size);
		            int val = beg;
		            if(ste==0) beg = end+1;
		            else beg = end - 1;
		            end = val;
		            ind--;
		            ste++;
		            ste %=2;
		            //System.out.println("al: "+al);
		        }
		    }
		    out.print(al.size());
		    for(int x:al) out.print(" "+x);
		    out.println();
		}
		out.flush();
	}
}