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 math
def last_char(last_index): # char is 0 or 1
	char = arr1[last_index]
	# print(char)
	reqd_first = char
	# print(reqd_first)
	# print(arr1)
	if (arr1[0] == reqd_first):
		return -1
	else:
		return 1
	# for idx in range(0,last_index):
	# 	if (arr1[idx] == 1-char):
	# 		return idx
	# return -2

def perform_operation(last_index):
	i = 0
	j = last_index
	while (i<=j):
		arr1[i], arr1[j] = 1-arr1[j], 1-arr1[i]
		i += 1
		j -= 1

num_cases = int(input())
for vvv in range(num_cases):
	n = int(input())
	arr1 = input().strip()
	arr1 = [int(elem) for elem in arr1]

	arr2 = input().strip()
	arr2 = [int(elem) for elem in arr2]

	# arr1 = list(map(int,input().strip().split()))
	# arr2 = list(map(int,input().strip().split()))
	ans = []
	for i in range(n-1,-1,-1):
		if (arr1[i] == arr2[i]):
			continue
		else:
			idx = last_char(i)
			# print(idx)
			if (idx == 1):
				ans.append(0)
				perform_operation(0)
				# print("i dont know")
			# elif (idx != 0):
			# 	ans.append(idx)
			# 	perform_operation(idx)
			ans.append(i)
			perform_operation(i)
	print(len(ans), end=' ')
	for elem in ans:
		print(elem+1, end=' ')
	print()



