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.
t=int(input())
for _ in range(t):
	# a=int(input())
	n=int(input())
	s1=list(input())
	s2=list(input())
	s12=s1[-1::-1]
	for i in range(len(s12)):
		if(s12[i]=="1"):
			s12[i]="0"
		else:
			s12[i]="1"
	s1inv=s12
	ans=[]
	ind=n
	flag=0
	for i in range(n):
		curr=n-i-1
		# print(curr,s1,s1inv,ans,flag)
		
		if(flag==0 and s2[curr]==s1[curr]):
			s1=s1[:-1]
			s1inv=s1inv[1:]
		elif(flag==1 and s2[curr]==s1inv[curr]):
			# print("kuma")
			s1=s1[1:]
			s1inv=s1inv[:-1]
		elif(flag==0 and s1[0]==s1[curr]):
			flag=1
			ans.append(curr)
			s1=s1[1:]
			s1inv=s1inv[:-1]
		elif(flag==1 and s1inv[0]==s1inv[curr]):
			# print("kuma2")
			flag=0
			ans.append(curr)
			# print(s1,s1inv)
			s1=s1[:-1]
			s1inv=s1inv[1:]
		elif(flag==0):
			flag=1
			ans.append(0)
			if(s1[0]=="0"):
				s1[0]="1"
				s1inv[-1]="0"
			else:
				s1[0]="0"
				s1inv[-1]="1"
			ans.append(curr)
			s1=s1[1:]
			s1inv=s1inv[:-1]
		else:
			flag=0
			ans.append(0)
			if(s1inv[0]=="0"):
				s1inv[0]="1"
				s1[-1]="0"
			else:
				s1inv[0]="0"
				s1[-1]="1"
			ans.append(curr)
			s1=s1[:-1]
			s1inv=s1inv[1:]
	for i in range(len(ans)):
		ans[i]+=1
		ans[i]=str(ans[i])
	if(len(ans)==0):
		print(0)
	else:
		print(len(ans)," ".join(ans))
