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:
def flip(arr_a, arr_b):
	ops = 0
	prefixlis = []
	lis_a = arr_a.copy()
	lis_b = arr_b.copy()
	while arr_a != arr_b:
		same = 0
		for i in range(-1, -len(arr_a)-1, -1):
			if arr_a[i] == arr_b[i]:
				same += 1
			else:
				break
		# print("same =", same)
		if same >= 1:
			arr_a = arr_a[:len(arr_a)-same]
			arr_b = arr_b[:len(arr_b)-same]
		if arr_a == []:
			# print("solution found")
			break
		# print("arr_a =", arr_a)
		# print("arr_b =", arr_b)
		if arr_a[0] == arr_b[-1]:
			prefixlis.append(1)
			prefixlis.append(len(arr_a))
			operation(arr_a, 1)
			operation(arr_a, len(arr_a))
			ops += 2
			if arr_a == arr_b:
				break
		else:
			prefixlis.append(len(arr_a))
			operation(arr_a, len(arr_a))
			ops += 1
			if arr_a == arr_b:
				break
	print(ops, end = " ")
	if ops == 0:
		return
	for i in range(ops):
		print(prefixlis[i], end = " ")

def operation(arr_a, prefix):
	for i in range(prefix):
		if arr_a[i] == "0":
			arr_a[i] = "1"
		else:
			arr_a[i] = "0"
	arr_a[:prefix] = arr_a[:prefix][::-1]
	# print("arr_a =", arr_a)


t = int(input())
for i in range(t):
	# ab = input().strip().split()
	n = int(input())
	# arr = list(map(int, input().strip().split()))
	a = input()
	b = input()
	arr_a = []
	for element in a:
		arr_a.append(element)
	arr_b = []
	for element in b:
		arr_b.append(element)
	# b = int(ab[1])
	# arr_a = list(map(int, input().strip().split()))
	# arr_b = list(map(int, input().strip().split()))
	if len(arr_a) == 1:
		if arr_a == arr_b:
			print(0)
		else:
			print(1, 1)
	else:
		flip(arr_a, arr_b)
		print()
"""
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
"""
"""
1
5
01011
11100
"""