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 CF1382C2 extends PrintWriter {
	CF1382C2() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1382C2 o = new CF1382C2(); o.main(); o.flush();
	}

	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			byte[] aa = sc.next().getBytes();
			byte[] bb = sc.next().getBytes();
			for (int i = 0; i < n; i++) {
				aa[i] -= '0';
				bb[i] -= '0';
			}
			int[] qu = new int[n * 2]; int cnt = 0, l = 0, r = n - 1, p = 0;
			for (int i = n - 1; i >= 0; i--) {
				int i_ = p == 0 ? l + i : r - i;
				int a = aa[i_] ^ p;
				int b = bb[i];
				if (a != b) {
					int j_ = p == 0 ? l : r;
					int c = aa[j_] ^ p;
					if (c != a) {
						qu[cnt++] = 1;
						aa[j_] ^= 1;
					}
					qu[cnt++] = i + 1;
					p ^= 1;
				}
				if (p == 0)
					r--;
				else
					l++;
			}
			print(cnt);
			for (int h = 0; h < cnt; h++)
				print(" " + qu[h]);
			println();
		}
	}
}