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.
import java.io.*;
import java.util.*;
import java.math.*;
public class A {
	public static void main(String[] args)throws IOException {
		FastIO sc = new FastIO(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int t = sc.nextInt();
		while(t-->0) {
			int n = sc.nextInt();
			String str = sc.next();
			int arr[] = new int[n];
			int nxt[] = new int[n];
			for(int i=0; i<n; ++i) {
				arr[i] = str.charAt(i)-'0';
			}
			str = sc.next();
			for(int i=0; i<n; ++i) {
				nxt[i] = str.charAt(i)-'0';
			}
			int perm[] = new int[n];
			for(int i=0; i<n; ++i) {
				perm[i] = (i%2==0) ? (n-i/2) : 2+i/2;
			}
			int save[] = new int[n];
			ArrayList<Integer> ans = new ArrayList<>();
			for(int i=n-1; i>=0; --i) {
				if(nxt[i]==arr[0]) {
					arr[0]^=1;
					ans.add(0);
				}
				ans.add(i);
				arr[0] = arr[perm[n-i-1]-1];
				if((n-1-i)%2==0)arr[0]^=1;
			}
			System.out.print(ans.size() + " ");
			for(int k: ans) {
				System.out.print((k+1) + " ");
			}
			System.out.println();
		}
		out.close();
	}
	static class FastIO {
		InputStream dis;
		byte[] buffer = new byte[1 << 17];
		int pointer = 0;
		public FastIO(String fileName) throws IOException {
			dis = new FileInputStream(fileName);
		}
		public FastIO(InputStream is) throws IOException {
			dis = is;
		}
		int nextInt() throws IOException {
			int ret = 0;
 
			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			boolean negative = false;
			if (b == '-') {
				negative = true;
				b = nextByte();
			}
			while (b >= '0' && b <= '9') {
				ret = 10 * ret + b - '0';
				b = nextByte();
			}
 
			return (negative) ? -ret : ret;
		}
		long nextLong() throws IOException {
			long ret = 0;
 
			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			boolean negative = false;
			if (b == '-') {
				negative = true;
				b = nextByte();
			}
			while (b >= '0' && b <= '9') {
				ret = 10 * ret + b - '0';
				b = nextByte();
			}
			return (negative) ? -ret : ret;
		}
		byte nextByte() throws IOException {
			if (pointer == buffer.length) {
				dis.read(buffer, 0, buffer.length);
				pointer = 0;
			}
			return buffer[pointer++];
		}
		String next() throws IOException {
			StringBuffer ret = new StringBuffer();
 
			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			while (b > ' ') {
				ret.appendCodePoint(b);
				b = nextByte();
			}
 
			return ret.toString();
		}
	}
}
