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 os import path
import sys
# mod = int(1e9 + 7)
# import re  # can use multiple splits
from math import ceil, floor,gcd,log
from collections import defaultdict , Counter
# from bisect import bisect_left, bisect_right
#popping from the end is less taxing,since you don't have to shift any elements
maxx = float('inf')

if (path.exists('input.txt')): 
	#------------------Sublime--------------------------------------#
	sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
	I = lambda :int(sys.stdin.buffer.readline())
	tup= lambda : map(int , sys.stdin.buffer.readline().split())
	lint = lambda :[int(x) for x in sys.stdin.buffer.readline().split()]
	S = lambda: sys.stdin.readline().replace('\n', '').strip()
	# def grid(r, c): return [lint() for i in range(r)]
	# def debug(*args, c=6): print('\033[3{}m'.format(c), *args, '\033[0m', file=sys.stderr)
	stpr = lambda x : sys.stdout.write(f'{x}' + '\n')
	star = lambda x: print(' '.join(map(str, x)))


else:
	#------------------PYPY FAst I/o--------------------------------#
	I = lambda :int(sys.stdin.buffer.readline())
	tup= lambda : map(int , sys.stdin.buffer.readline().split())
	lint = lambda :[int(x) for x in sys.stdin.buffer.readline().split()]
	S = lambda: sys.stdin.readline().replace('\n', '').strip()
	# def grid(r, c): return [lint() for i in range(r)]
	stpr = lambda x : sys.stdout.write(f'{x}' + '\n')
	star = lambda x: print(' '.join(map(str, x)))

for _ in range(int(input())):
	n = int(input())
	a = input().strip()
	b = input().strip()
	if a == b:
		print(0)
		continue
	pre =[]
	for i in range(n):
		if a[i] != b[i]:
			if i > 0 :
				pre.append(i+1)
			pre.append(1)
			if i > 0 :
				pre.append(i+1)
	print(len(pre) , *pre)