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.
def solve(s):
	new = ""
	for i in s:
		if i=="1":
			new += "0"
		else:
			new += "1"
	return new[::-1]

for nt in range(int(input())):
	n = int(input())
	a = input()
	b = input()
	ans = []
	i = 1
	count = 0
	if a[0]!=b[0]:
		ans.append(1)
	while i<n:
		if i+1<n:
			if a[i]!=b[i] and a[i+1]!=b[i+1]:
				if a[i]==a[i+1]:
					ans.append(i+2)
					ans.append(2)
					ans.append(i+2)
					i += 2
				else:
					if i+2>=n or a[i+2]==b[i+2]:
						ans.append(i+2)
						ans.append(1)
						ans.append(2)
						ans.append(1)
						ans.append(i+2)
						i += 3
						continue
					if a[i+2]==a[i]:
						ans.append(i+3)
						ans.append(3)
						ans.append(i+3)
						i += 3
					else:
						ans.append(i+3)
						ans.append(1)
						ans.append(3)
						ans.append(1)
						ans.append(i+3)
						i += 3
			elif a[i]==b[i] and a[i+1]!=b[i+1]:
				i += 1
			elif a[i]!=b[i] and a[i+1]==b[i+1]:
				ans.append(i+1)
				ans.append(1)
				ans.append(i+1)
				i += 2
			else:
				i += 2
		else:
			if a[i]!=b[i]:
				ans.append(i+1)
				ans.append(1)
				ans.append(i+1)
			break

	


	print (len(ans),end = " ")
	print (*ans)



# def solve(s):
# 	new = ""
# 	for i in s:
# 		if i=="1":
# 			new += "0"
# 		else:
# 			new += "1"
# 	return new[::-1]

# for nt in range(int(input())):
# 	n = int(input())
# 	a = input()
# 	b = input()
# 	ans = []
# 	for i in range(n-1,-1,-1):
# 		print (a,i,a[i])
# 		if a[i]!=b[i]:
# 			if a[i]==a[0]:
# 				ans.append(i+1)
# 				a = solve(a[0:i+1]) + a[i+1:]
# 			else:
# 				ans.append(1)
# 				ans.append(i+1)
# 				a = solve(a[i]+a[1:i+1]) + a[i+1:]

# 	print (len(ans),end = " ")
# 	print (*ans)