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 collections import defaultdict

rr = lambda: input()
rri = lambda: int(input())
rrm = lambda: list(map(int, input().split()))

INF=float('inf')

def verify(steps, A,B):
	print("verify")
	for j, s in enumerate(steps):
		s = int(s)
		a = A[:s]
		for i in range(s):
			a[i] ^= 1
		a = a[::-1]
		for i in range(s):
			A[i] = a[i]
		#print(A, "step", j, s)
	for v1,v2 in zip(A,B):
		if v1 != v2:
			return False
	return True

def solve(N,A,B):
	if A == B:
		print(0)
		return
	# at most 3n operations
	A = list(map(int, list(A)))
	B = list(map(int, list(B)))
	C = A.copy()

	# operation, flips all bits and reverses
	lo = 0 # lo is always 0
	hi = len(A)-1

	steps = []
	while lo <= hi:
		if A[hi] == B[hi]:
			# nothing, no flips
			pass
		else:
			steps.append(str(hi+1))
			steps.append(str(lo+1))
			steps.append(str(hi+1))
			# A[hi] should be fixed to B[hi], 3 flips
		hi -= 1
	# verify
	#print(verify(steps,C,B))

	print(len(steps,), ' '.join(steps))

t = rri()
for _ in range(t):
	N=rri()
	A=rr()
	B=rr()
	solve(N,A,B)
