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.

g=int(input())
for _ in range(g):
	n=int(input())
	A=input()
	B=input()
	count=0
	C=[]
	for i in range(n-1,-1,-1):
		#print(i)
		if(B[i]!=A[i]):
			if(B[i]!=A[0]):
				C.append(i+1)
				mystr=""
				for j in range(0,i+1):
					if(A[j]=='0'):
						mystr=str(1)+mystr
					else:
						mystr=str(0)+mystr

				A=mystr+A[i+1:]

				count+=1
				#print("After ",count,"A=",A)
			else:
				C.append(1)
				A=str(1-int(A[0]))+A[1:]
				C.append(i+1)
				E=list(A[:i+1])
				mystr=""
				for j in range(0,i+1):
					if(A[j]=='0'):
						mystr=str(1)+mystr
					else:
						mystr=str(0)+mystr

				A=mystr+A[i+1:]
				count+=2
				#print(mystr)
				#print("After ",count,"A=",A)
	if(count==0):
		print(count)
	else:
		print(count,end=" ")
		for i in range(0,count):
			if(i==count-1):
				print(C[i])
			else:
				print(C[i],end=" ")





  		
  

