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 PYTHON):

Solution:


# template by 3xC and starkizard.
# contributors: 
    
#####################################################################################
from __future__ import division, print_function 
    
import sys
import os
from collections import Counter, deque, defaultdict
import itertools
import math
import io
    
    
"""Uncomment modules according to your need"""
    
    
# from bisect import bisect_left, bisect_right, insort
# from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
# from copy import deepcopy
# from decimal import *
# import re
# import operator
    
#####################################################################################
# this enables you to write python3 code with PyPy2 (Python 2)
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange
    
    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip
    
#####################################################################################
    
"""value of mod"""
MOD = 998244353
mod = 10**9 + 7
    
"""Uncomment next 4 lines if doing huge recursion"""
# import threading
# threading.stack_size(1<<27)
# sys.setrecursionlimit(10000
    
def prepare_factorial(mod=mod):
    """ returns two lists, factorial and inverse factorial modulo argument by default 10**9 +7 """
    # Comment code out when you don't need inverse factorial or vice versa
    fact = [1]
    for i in range(1, 200005):
        fact.append((fact[-1] * i) % mod)
    
    ifact = [0] * 200005
    ifact[200004] = pow(fact[200004], mod - 2, mod)
    for i in range(200004, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod
    
    return fact, ifact
    
def modinv(n, p):
    """ returns N inverse modulo p """
    return pow(n, p - 2, p)
    
def ncr(n, r,  fact, ifact):
    """ takes 4 arguments: n , r and factorial and inverse factorial lists"""
    t = (fact[n] * (ifact[r]*ifact[n-r]) % MOD)% MOD
    return t
    
    
def get_n(Sum):    
    """this function returns the maximum n for which Summation(n) <= Sum"""
    ans = (-1 + sqrt(1 + 8*Sum))//2
    return ans
    
def sieve(n):
    """ returns a list of prime numbers till n """
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r
    
    
    
def divs(n, start=1):
    """ returns a list of all divisors till n """
    divisors = []
    #rdivisors=[]
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors
    
    
    
def divn(N):
    """ returns the number of divisors """
    divs_number = 1
    factorCounter=1
    tempList=[]
    
    while N%2==0:
        factorCounter+=1
        N=N//2
    tempList.append(factorCounter)
    
    i=3
    
    while i*i<=N:
        factorCounter=1
        while N%i==0:
            factorCounter+=1
            N=N//i
        if factorCounter!=1:
            tempList.append(factorCounter)
        i+=1
    
    if N>2:
        factorCounter=2
        tempList.append(factorCounter)
    
    divisorCount=1
    for i in tempList:
        divisorCount*=i
    return(divisorCount)
    
    
    
def lrfind(d, x, default=-1):
    """ Takes 2 arguments an iterable and an element. returns a tuple (firstoccurence,lastoccurence) -1 if not found """
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return default, default
    else:
        return left, right
    
def gcd(x, y): # math.gcd is slower
    """ returns greatest common divisor of x and y """
    while y:
        x, y = y, x % y
    return x
    
def check_sorted(a):
    ''' returns True/False '''
    for i in range(len(a)-1):
        if a[i]>a[i+1]:
            return False
    return True
def ceil(n, k=1): return n // k + (n % k != 0) #returns math.ceil but protecting against floating inconsistencies
def input(): return sys.stdin.readline().strip()
def ii(): return int(input()) #inputs integer
def mi(): return map(int, input().split()) # inputting space seperated variables for example x,y,z
def li(): return list(map(int, input().split())) #inputting a space seperated list of integers
def lw(): return input().split() #inputting a space seperated list of strings
def lcm(a, b): return abs(a * b) // gcd(a, b) #returns LCM of two arguments
def prr(a, sep=' ', end='\n'): print(sep.join(map(str, a)), end=end) #For printing an iterable with seperator sep as optional second argument (default : " "), ending character (default: "\n") as optional third
def dd(): return defaultdict(int) #returns a dictionary with values defaulted to 0
def ddl(): return defaultdict(list) #returns a dictionary with values defaulted to []
def write(s): return sys.stdout.write(s)
    
    
    
    
    
    
    
    
    
    
    
    
###################################################################
    
def main():
    #CODE GOES HERE:
    for t in range(ii()):
        n=ii()
        a=list(input())
        b=list(input())
        answer=[]
        r=n-1
        while(r>-1):
            if a[r]==b[r]:
                r-=1
            else:
                if a[0]==a[r]:
                    for i in range(r+1):
                        if a[i]=="0":
                            a[i]="1"
                        else:
                            a[i]="0"
                    a[:r+1]=a[:r+1][::-1]   
                    answer.append(r+1)
                    r-=1
                else:
                    answer.append(1)
                    a[0]=["0","1"][a[0]=="0"]
        print(len(answer),end=" ")
        prr(answer)

        
            
    
    
""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""
    
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
    
import os, sys
from io import IOBase, BytesIO
    
BUFSIZE = 8192
    
class FastIO(BytesIO):
    newlines = 0
    
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
    
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s
    
    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()
    
    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
    
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
    
    
class IOWrapper(IOBase):
    
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')
    
    
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
    
""" main function"""
    
if __name__ == '__main__':
    main()
    # threading.Thread(target=main).start()