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.
for i in range(int(input())):
	n= int(input())
	a = input()#first string
	b = input()#second string
	d = []
	for c in range(1,n+1):
		if a == b:
			print (len(d),*d)
			break
			
		if int(a[-c]) == int(b[-c]):
#compare last element of both string
			continue
		else:
			if int(a[0]) == int(b[-c]):
#compare first element a  with last element of b 
				d.append(1)#take prefix 1
				
				if int(a[0]) == 0:
					a=list(a)
					a[0] = "1"
					a = "".join(a)
				# there just change the first element of a		
				else:
					a = list(a)
					a[0] = "0"
					a = "".join(a)
					
					
			if a == b:
				print(len(d),*d)
				break	
					
			d.append(n-c+1)
	#take prefix length of a where "a" element can't equal "b"  .ex:  a = 0(1)1 , b = 1(0)1  here bracket portion didn't match so take prefix 2
	 
			k = a[:n-c+1].replace("1","k")
			k = k.replace("0","m")
			x = k.replace("k","0")
			x = x.replace("m","1")
			if len(x) == n:
				a = x[::-1]
			else:
				a = x[::-1] + a[-(c-1):]
				
	# here just change the value of prefix and reverse it...ex : a = 101 take prefix 2 as 10
	# k = k0 just change 1 to k
	# k = km change 0 to m
	# x = 0m change k to 0
	# x = 01 change m to 1
	# then prefix reverse and it turns to a again		
			if a == b:
				print (len(d),*d)
				break	