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.
from sys import stdin
g   = lambda : stdin.readline().strip()
gl  = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]

from collections  import deque
removeLast = lambda a, rev : a.popleft() if rev else a.pop()
getLast = lambda a, rev, inv : a[0]^inv if rev else a[-1]^inv
getFirst = lambda a, rev, inv : a[-1]^inv if rev else a[0]^inv


t, = gil()
for _ in range(t):
	inv, rev = 0 , 0 
	n, = gil()
	a, b = deque(gbs()), gbs()
	r = n - 1
	ans = []
	while r != 0 :
		if getLast(a, rev, inv) != b[r] :
			if getLast(a, rev, inv) != getFirst(a, rev, inv) :
				if rev : 
					a[-1] ^= inv^1
				else :
					a[0] ^= inv^1

				ans.append(1)
			ans.append(r+1)
			inv ^= 1
			rev ^= 1
		removeLast(a, rev)
		r -= 1

	if a[0]^inv != b[0]:
		ans.append(1)

	print(len(ans), end = " ")
	for a in ans :
		print(a, end = " ")
	print()