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.
import sys
# from collections import deque
# from collections import Counter
# from math import sqrt
# from math import log
# from math import ceil
# from bisect import bisect_left, bisect_right

# alpha=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
# mod=10**9+7
# mod=998244353

# def BinarySearch(a,x): 
# 	i=bisect_left(a,x) 
# 	if(i!=len(a) and a[i]==x): 
# 		return i 
# 	else: 
# 		return -1

# def sieve(n): 
# 	prime=[True for i in range(n+1)]
# 	p=2
# 	while(p*p<=n): 
# 		if (prime[p]==True): 
# 			for i in range(p*p,n+1,p): 
# 				prime[i]=False
# 		p+=1
# 	prime[0]=False
# 	prime[1]=False
# 	s=set()
# 	for i in range(len(prime)):
# 		if(prime[i]):
# 		s.add(i)
# 	return s

# def gcd(a, b):
# 	if(a==0):
# 		return b 
# 	return gcd(b%a,a)

fast_reader=sys.stdin.readline
fast_writer=sys.stdout.write

def input():
	return fast_reader().strip()

def print(*argv):
	fast_writer(' '.join((str(i)) for i in argv))
	fast_writer('\n')

#____________________________________________________________________________________________________________________________________

for _ in range(int(input())):
	n=int(input())
	a=list(input())
	b=list(input())
	if(a==b):
		print(0)
		continue
	l=a[:]
	curr=0
	m=[]
	for i in range(n):
		if(l[i]=='0'):
			if(curr!=0):
				if(i==curr):
					m.append(i)
				else:
					m.append(i-curr)
					m.append(i)
			curr=0
		else:
			curr+=1
	if(curr!=0):
		if(n==curr):
			m.append(n)
		else:
			m.append(n-curr)
			m.append(n)
	#print(m)
	temp=['0' for i in range(n)]
	i=n-1
	while(i>=0):
		if(temp[i]!=b[i]):
			j=i
			curr=0
			while(j>=0):
				if(temp[j]!=b[j]):
					curr+=1
					j-=1
				else:
					break
			if(j==-1):
				m.append(curr)
			else:
				m.append(i+1)
				m.append(j+1)
			i=j
		i-=1
	print(len(m),*m)