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.util.*;
public class prefixflip {

	public static void main(String[] args){
		// TODO Auto-generated method stub
		Scanner inputs = new Scanner(System.in);
		int loo = inputs.nextInt();
		for(int p=0;p<loo;p++){
			int n = inputs.nextInt();
			String a = inputs.next();
			String b = inputs.next();
			int ind = a.length()-1;
			int num = 0;
			StringBuilder ans = new StringBuilder();
			while(ind>0) {
				if(a.charAt(ind)!=b.charAt(ind)) {
					//then we are good
					if(a.charAt(0)!=b.charAt(ind)) {
						a = flip(a,ind);
						num++;
						ans.append((ind+1)+" ");
					}
					else {
						a = flip(a,0);
						num++;
						ans.append("1 ");
						a = flip(a,ind);
						num++;
						ans.append((ind+1)+" ");
					}
				}
				ind--;
			}
			if(a.charAt(0)!=b.charAt(0)) {
				num++;
				ans.append("1 ");
			}
			if(ans.length()==0) {
				System.out.println(0);
			}
			else {
				ans.deleteCharAt(ans.length()-1);
				System.out.println(num+" "+ans);
			}
		}

	}
	public static String flip(String s,int j) {
		String sub = "";
		for(int p=j;p>=0;p--) {
			if(s.charAt(p)=='1') {
				sub+='0';
			}
			else {
				sub+='1';
			}
		}
		return sub+s.substring(j+1);
	}

}