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 PYTHON3):

Solution:

import sys
sys.setrecursionlimit(10000)
# default is 1000 in python


t = int(input())
# t = 1

for _ in range(t):
	n = int(input())

	a = list(input())
	b = list(input())
	prefs = [0]

	reverse = False
	invert = False

	la = 0
	ra = n-1
	lb = 0
	rb = n-1

	while ra-la+1 > 0:
		# print(a)
		# print(b)
		if reverse is False:

			if (a[ra] == b[rb] and invert is False) or (a[ra] != b[rb] and invert is True):
				# a = a[:-1]
				ra -= 1
				rb -= 1
				continue

			if (a[la] != b[rb] and invert is False) or (a[la] == b[rb] and invert is True):
				prefs.append(ra-la+1)
				invert = not invert
				reverse = not reverse
				# c = [str(1-int(i)) for i in a]
				# c.reverse()
				# a = c.copy()
			else:
				prefs.append(1)
				prefs.append(ra-la+1)
				reverse = not reverse
				invert = not invert
				la += 1
				rb -= 1 

		else:
			if (a[la] == b[rb] and invert is False) or (a[la] != b[rb] and invert is True):
				la += 1
				rb -= 1
				continue

			if (a[ra] != b[rb] and invert is False) or (a[ra] == b[rb] and invert is True):
				prefs.append(ra-la+1)
				invert = not invert
				reverse = not reverse

			else:
				prefs.append(1)
				prefs.append(ra-la+1)
				reverse = not reverse
				invert = not invert
				ra -= 1
				rb -= 1

	prefs[0] = len(prefs) - 1
	print(*prefs)


# try:
	# raise Exception
# except:
	# print("-1")


# from collections import OrderedDict 
# mydict = OrderedDict() 


# thenos.sort(key=lambda x: x[2], reverse=True)

# int(math.log(max(numbers)+1,2))

# 2**3 (power)

# a,t = (list(x) for x in zip(*sorted(zip(a, t))))

# to copy lists use .copy()

# pow(p, si, 1000000007) for modular exponentiation

# my_dict.pop('key', None)
# This will return my_dict[key] if key exists in the dictionary, and None otherwise.



# bin(int('010101', 2))