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 math
from decimal import Decimal
import heapq
import copy
import heapq
from collections import deque
def na():
	n = int(input())
	b = [int(x) for x in input().split()]
	return n,b
 
	
def nab():
	n = int(input())
	b = [int(x) for x in input().split()]
	c = [int(x) for x in input().split()]
	return n,b,c
 
		
def dv():
	n, m = map(int, input().split())
	return n,m
 
 
def da():
	n, m = map(int, input().split())
	a = list(map(int, input().split()))
	return n,m, a 
 
 
def dva():
	
	n, m = map(int, input().split())
	a = [int(x) for x in input().split()]
	b = [int(x) for x in input().split()]
	return n,m,b
 
 
def eratosthenes(n): 
	sieve = list(range(n + 1))
	for i in sieve:
		if i > 1:
			for j in range(i + i, len(sieve), i):
				sieve[j] = 0
	return sorted(set(sieve))
 
 
def lol(lst,k):
	k=k%len(lst)
	ret=[0]*len(lst)
	for i in range(len(lst)):
		if i+k<len(lst) and i+k>=0:
			ret[i]=lst[i+k]
		if i+k>=len(lst):
			ret[i]=lst[i+k-len(lst)]
		if i+k<0:
			ret[i]=lst[i+k+len(lst)]
	return(ret)
def nm():
	n = int(input())
	b = [int(x) for x in input().split()]
	m = int(input())
	c = [int(x) for x in input().split()]
	return n,b,m,c
 
 
def dvs():
	n = int(input())
	m = int(input())
	return n, m 
 
 
def fact(n):
	tc = []
	ans = {}
	d = 2
	while d * d <= n:
		if n % d == 0:
			tc.append(d)
			n //= d
		else:
			d += 1
	if n > 1:
		tc.append(n)
	for i in tc:
		ans[i] = ans.get(i, 0) + 1
	return ans


def to_zero(n, s):
	lst = s[0]
	ans = []
	for i in range(n):
		if lst == s[i]:
			continue
		else:
			if lst == '0':
				lst = '1'
				ans.append(i)
			else:
				lst = '0'
				ans.append(i)
	if lst == '1':
		ans.append(n)
	return ans



for _ in range(int(input())):
	n = int(input())
	s = input()
	t = input()
	ans = to_zero(n, s) + to_zero(n, t)[::-1]
	ans = [len(ans)] + ans
	print(*ans)
