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.util.*;

public class Sample
{

	static int[][][] dp;
	static int[] are;

	public static void main(String[] args) throws InterruptedException {
		Thread t=new Thread(null, null,"", 1<<28) {
			public void run() {
				m();
			}
		};
		t.start();
		t.join();
	}

	public static void m()
	{
		FastScanner in=new FastScanner();
		int t = in.nextInt();
		while(t-->0)
		{
			int n = in.nextInt();
			char[] a = in.next().toCharArray();
			char[] b = in.next().toCharArray();
			//
			ArrayList<Integer> ansA = new ArrayList<>();
			ArrayList<Integer> ansB = new ArrayList<>();
			for(int i=1; i<n; i++) if(a[i]!=a[i-1]) ansA.add(i);
			if(a[n-1]=='1') ansA.add(n);
			for(int i=1; i<n; i++) if(b[i]!=b[i-1]) ansB.add(i);
			if(b[n-1]=='1') ansB.add(n);
			//
			System.out.print((ansA.size()+ansB.size())+" ");
			for(int i=0; i<ansA.size(); i++) System.out.print(ansA.get(i)+" ");
			for(int i=ansB.size()-1; i>=0; i--) System.out.print(ansB.get(i)+" ");
			System.out.println();
		}
	}

/*
1
5
01101
10101

*/

	static void longestsub(int[] str1, int[] str2)
	{
		int n = str1.length;
		int m = str2.length;
		//
		int[][] dp = new int[n+1][m+1];
		//
		for(int i=0; i<=n; i++)
		{
			for(int j=0; j<=m; j++)
			{
				if(i==0 || j==0) {dp[i][j]=0; continue;}
				//
				if(str1[i-1]==str2[j-1]) dp[i][j]=dp[i-1][j-1]+1;
				else dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
			}
		}
		//
		int length = dp[n][m];
		//
		String ans = "";
		//
		int i=n, j=m;
		while(true)
		{
			if(i==0 || j==0) break;
			//
			if(dp[i][j]==Math.max(dp[i-1][j], dp[i][j-1]))
			{
				if(dp[i-1][j]>dp[i][j-1]) i--;
				else j--;
			}
			else
			{
				ans = str1[i-1] + " " + ans;
				i--; j--;
			}
		}
		//
		System.out.println(length>0?"YES"+'\n'+"1 "+ans.substring(0,1):"NO");
	}

	static void print(String[] are)
	{
		for(int i=0; i<are.length; i++)
			System.out.print(are[i]);
		System.out.println();
	}

	static void sort(int[] a)
	{
		ArrayList<Integer> list=new ArrayList<>();
		for (int i:a) list.add(i);
		Collections.sort(list);
		for (int i=0; i<a.length; i++) a[i]=list.get(i);
	}

	static class FastScanner
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");

		public String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();

		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}

}