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.

/*
                       _oo0oo_
                      o8888888o
                      88" . "88
                      (| -_- |)
                      0\  =  /0
                    ___/`---'\___
                  .' \\|     |// '.
                 / \\|||  :  |||// \
                / _||||| -:- |||||- \
               |   | \\\  -  /// |   |
               | \_|  ''\---/''  |_/ |
               \  .-\__  '-'  ___/-. /
             ___'. .'  /--.--\  `. .'___
          ."" '<  `.___\_<|>_/___.' >' "".
         | | :  `- \`.;`\ _ /`;.`/ - ` : | |
         \  \ `_.   \_ __\ /__ _/   .-` /  /
     =====`-.____`.___ \_____/___.-`___.-'=====
                       `=---='
 
*/
import java.util.function.Consumer;
import java.util.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.io.*;
import java.lang.Math.*;
 public class KickStart2020{
	static 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());}
        float nextFloat() {return Float.parseFloat(next());}
        String nextLine() {
            String str = "";
            try {str = br.readLine();}
            catch (IOException e) { e.printStackTrace();}
            return str; }}
	static String reverseOfString(String a) {
		StringBuilder ssd = new StringBuilder();
		for(int i = a.length() - 1; i >= 0; i--) {
			ssd.append(a.charAt(i));
		}
		return ssd.toString();
	}
	static char[] reverseOfChar(char a[]) {
		char b[] = new char[a.length];
		int j = 0;
		for(int i = a.length - 1; i >= 0; i--) {
			b[i] = a[j];
			j++;
		}
		return b;
	}
	static boolean isPalindrome(char a[]) {
		boolean hachu = true;
		for(int i = 0; i <= a.length / 2; i++) {
			if(a[i] != a[a.length - 1 - i]) {
				hachu = false;
				break;
			}
		}
		return hachu;
	} 
	static long gcd(long a, long b)
	{
	    if (b == 0)
	        return a;
	    return gcd(b, a % b);
	     
	}
	static boolean check(String a) {
		boolean hachu = true;
		for(int i = 0; i < a.length(); i++) {
			if(a.charAt(0) != a.charAt(i)) {hachu = false; break;}
		}
		return hachu;
	}
	static void move(char a[], int i) {
		int j = 0;
		int temp1 = i;
		while(j < i) {
			char temp = a[i];
			a[i] = a[j];
			a[j] = temp;
			j++;
			i--;
		}
		for(int k = 0; k <= temp1; k++) {
			a[k] = (a[k] == '0') ? '1' : '0';
		}
	}
		public static void main(String[] args) throws Exception{
FastReader sc = new FastReader();
int t = sc.nextInt();
while(t-- > 0) {
	int n = sc.nextInt();
	char a[] = sc.next().toCharArray();
	char b[] = sc.next().toCharArray();
	ArrayList<Integer> ss = new ArrayList<>();
	for(int i = n - 1; i >= 0; i--) {
		if(a[i] != b[i]) {
			if(a[i] == a[0]) {
				ss.add(i + 1);
				move(a, i);
			}
			else {
				ss.add(1);
				a[0] = (a[0] == '0') ? '1' : '0';
				ss.add(i + 1);
				move(a, i);
			}
		}
	}
	System.out.print(ss.size() + " ");
	for(int e: ss) System.out.print(e + " ");
	System.out.println();
}
}
}