Problem: 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.

Solution (in PYTHON3):

Solution:
from sys import stdin,stdout                           #
import math                                            #
import heapq                                           #
                                                       #
t = 1                                                  #
def aint():                                            #
	return int(input().strip())                        #
def lint():                                            #
	return list(map(int,input().split()))              #
def fint():                                            #
	return list(map(int,stdin.readline().split()))     #
                                                       #	
########################################################

def main():
	n=aint()
	s1=input()
	s2=input()
	cnt=0
	ans=[]
	s,e=0,n-1
	for i in range(n-1,-1,-1):
		if cnt%2==0:
			b1,b2=s1[s]=='1',s2[i]=='1'
			s+=1
		else:
			b1,b2=s1[e]=='0',s2[i]=='1'
			e-=1
		if b1==b2:
			ans.append(1)
		ans.append(i+1)
		cnt+=1
	print(len(ans),*ans)

t=int(input())

########################################################
for i in range(t):                                     #
	#print("Case #"+str(i+1)+":",end=" ")		       #
	main()                                             #