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:
#div1
import sys

def input():
	return sys.stdin.readline().rstrip()

def input_split():
	return [int(i) for i in input().split()]

testCases = int(input())
answers = []

def update_zero(arr, diff, num_flips):
	if num_flips%2 == 0:
		i = -diff
		arr[i] = 1 - arr[i]
		# ans = arr[i]
	else:
		i = diff
		# ans =  1 - arr[i]
		arr[i] = 1 -arr[i]

def fetch_val(arr, index, diff, num_flips):
	# if index == 0:
		# return arr[]
	if num_flips%2 == 0:
		i = -diff + index
		ans = arr[i]
	else:
		i = diff - index
		ans =  1 - arr[i]

	# print('fetching index {} of arr, it is {}'.format(index, ans))
	return ans
	# if index == 0:

def print_arr(arr, diff, num_flips, ans_till_now):
	# temp = [fetch_val(arr, i, diff, num_flips) for i in range(len(arr))]
	temp = [fetch_val(arr, i, diff, num_flips) for i in range(ans_till_now[-1])] 
	temp = temp + arr[ans_till_now[-1]:]
	print('array is ')
	print(*temp, sep = " ")

for _ in range(testCases):
	#take input
	n = int(input())
	arr = [int(i) for i in input()]
	barr = [int(i) for i in input()]
	# barr = input_split()
	ans = []
	num_flips = 0
	# one_flips = 0
	diff = 0
	for i in range(n):
		index = n-1 -i
		desired = barr[index]

		if fetch_val(arr, index, diff, num_flips) == desired:
			# print("")
			continue
		else:
			target = fetch_val(arr, 0, diff, num_flips)
			# if one_flips%2 == 1:
			# 	target = 1 - target

			if target == 1 - desired:
				ans.append(index+1) #check?
				num_flips += 1
				diff = index - diff
			else:
				ans.append(1)
				update_zero(arr, diff, num_flips)
				ans.append(index+1)
				num_flips += 1
				diff = index - diff
				# one_flips += 1
				# arr[0] = 1 - arr[0]
		# print('ops till now are {}'.format(ans))
		# print_arr(arr, diff, num_flips, ans)

			# temp = [ 1- arr[i] for i in range(index + 1)]
			# temp.reverse()
			# arr = temp + arr[index + 1:]

	answers.append(ans)

for ans in answers:
	print(len(ans), end = ' ')
	print(*ans, sep = ' ')