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 os , sys,time, collections , math , pprint , itertools as it , operator as op , bisect as bs ,functools as fn
maxx , localsys , mod = 1<<60, 0 , int(1e9 + 7) 
nCr = lambda n, r: reduce(mul, range(n - r + 1, n + 1), 1) // factorial(r)
ceil = lambda n , x: (n+x -1 )//x 
osi, oso = '/home/priyanshu/Documents/cp/input.txt','/home/priyanshu/Documents/cp/output.txt'
if os.path.exists(osi):
	sys.stdin  = open(osi, 'r') ; sys.stdout = open(oso, 'w')
	
input = sys.stdin.readline
from functools import reduce
def maps():return map(int , input().split())

#THINK ABOUT THE EDGE CASES ..........

"""for _ in range(int(input())):
	n = int(input()) ; a = list(maps())
	pos = n- 1
	while pos>0 and a[pos-1]>=a[pos]:pos-=1
	while pos>0 and a[pos-1]<=a[pos]:pos-=1
	print(pos)

longest suffix which is sorted either increasing or decreasing doesn't matter
"""

"""

for _ in range(int(input())):
	n = int (input())
	t = [98]*51
	for i in [0] + list(maps()):
		t[i]^=1
		print(''.join(map(chr,t)))

"""


"""
for _ in range(int(input())):
	n = int(input()) ; a = list(maps())
	cnt = a.count(1)
	if cnt == n:
		print('First') if n%2 else print('Second')
	else:
		k=0
		while k <n and a[k]==1:
			k+=1
		print('First' if k%2==0 else 'Second')

"""




for _ in range(int(input())):
	n = int(input()); a = list(map(int,input().rstrip('\n'))) ; b = list(map(int,input().rstrip('\n')))
	#a --> b
	ans =[]
	for i in range(n):
		if a[i] != b[i]:
			ans.append(i+1)
			ans.append(1)
			ans.append(i+1)

	print(len(ans),*ans)