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 PrefixFlip {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(br.readLine());

		for (int t = 0; t < T; t++) {
			int N = Integer.parseInt(br.readLine());
			String temp = br.readLine();
			int[] a = new int[N];
			for (int i = 0; i < N; i++)
				a[i] = temp.charAt(i) - 48;
			String b = br.readLine();

			Queue<Integer> q = new LinkedList<Integer>();
			boolean flipped = false;
			int p1 = 0; // first pointer
			int p2 = N - 1; // second pointer
			for (int i = N - 1; i >= 0; i--) {
				int bitA = flipped ? 1 - a[p2] : a[p2];
				int bitB = b.charAt(i) - 48;
				if (bitA == bitB) {
					if(flipped) 
						p2++;
					else
						p2--;
					continue;
				}
				int bit1 = flipped ? 1 - a[p1] : a[p1];
				if(bit1 == bitB)
					q.offer(1);
				q.offer(i+1);
				if(!flipped) {
					int prev = p2;
					p2 = p1+1;
					p1 = prev;
				} else {
					int prev = p1;
					p1 = p2;
					p2 = prev-1;
				}
				flipped = !flipped;
			}
			System.out.print(q.size() + " ");
			while(!q.isEmpty())
				System.out.print(q.poll() + " ");
			System.out.println();
		}
	}
}