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:
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from decimal import *
from io import BytesIO, IOBase
from itertools import *
from collections import *

# sys.setrecursionlimit(10**5)
M = 10 ** 9 + 7

# print(math.factorial(5))
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)

# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        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")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")


def inp(): return sys.stdin.readline().rstrip("\r\n")  # for fast input


def out(var): sys.stdout.write(str(var))  # for fast output, always take string


def lis(): return list(map(int, inp().split()))


def stringlis(): return list(map(str, inp().split()))


def sep(): return map(int, inp().split())


def strsep(): return map(str, inp().split())


def fsep(): return map(float, inp().split())


def inpu(): return int(inp())


# -----------------------------------------------------------------

def regularbracket(t):
    p = 0
    for i in t:
        if i == "(":
            p += 1
        else:
            p -= 1
        if p < 0:
            return False
    else:
        if p > 0:
            return False
        else:
            return True


# -------------------------------------------------
def binarySearchCount(arr, n, key):
    left = 0
    right = n - 1

    count = 0

    while (left <= right):
        mid = int((right + left) / 2)

        # Check if middle element is
        # less than or equal to key
        if (arr[mid] <= key):
            count = mid + 1
            left = mid + 1
        # If key is smaller, ignore right half
        else:
            right = mid - 1
    return count


# ------------------------------reverse string(pallindrome)
def reverse1(string):
    pp = ""
    for i in string[::-1]:
        pp += i
    if pp == string:
        return True
    return False


# --------------------------------reverse list(paindrome)
def reverse2(list1):
    l = []
    for i in list1[::-1]:
        l.append(i)
    if l == list1:
        return True
    return False


def mex(list1):
    # list1 = sorted(list1)
    p = max(list1) + 1
    for i in range(len(list1)):
        if list1[i] != i:
            p = i
            break
    return p


def sumofdigits(n):
    n = str(n)
    s1 = 0
    for i in n:
        s1 += int(i)
    return s1


def perfect_square(n):
    s = math.sqrt(n)
    if s == int(s):
        return True
    return False


# -----------------------------roman
def roman_number(x):
    if x > 15999:
        return
    value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
    symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
    roman = ""
    i = 0
    while x > 0:
        div = x // value[i]
        x = x % value[i]
        while div:
            roman += symbol[i]
            div -= 1
        i += 1
    return roman


def soretd(s):
    for i in range(1, len(s)):
        if s[i - 1] > s[i]:
            return False
    return True


# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
    ct = 0
    for i in range(2, h + 1, 2):
        for j in range(2, w + 1, 2):
            ct += (h - i + 1) * (w - j + 1)
    return ct


def countrhombi2(h, w):
    return ((h * h) // 4) * ((w * w) // 4)


# ---------------------------------
def binpow(a, b):
    if b == 0:
        return 1
    else:
        res = binpow(a, b // 2)
    if b % 2 != 0:
        return res * res * a
    else:
        return res * res


# -------------------------------------------------------
def binpowmodulus(a, b, m):
    a %= m
    res = 1
    while (b > 0):
        if (b & 1):
            res = res * a % m
        a = a * a % m
        b >>= 1
    return res


# -------------------------------------------------------------
def coprime_to_n(n):
    result = n
    i = 2
    while (i * i <= n):
        if (n % i == 0):
            while (n % i == 0):
                n //= i
            result -= result // i
        i += 1
    if (n > 1):
        result -= result // n
    return result


# -------------------prime
def prime(x):
    if x == 1:
        return False
    else:
        for i in range(2, int(math.sqrt(x)) + 1):
            # print(x)
            if (x % i == 0):
                return False
        else:
            return True

def luckynumwithequalnumberoffourandseven(x,n,a):
    if x >= n and str(x).count("4") == str(x).count("7"):
        a.append(x)
    else:
        if x < 1e12:
            luckynumwithequalnumberoffourandseven(x * 10 + 4,n,a)
            luckynumwithequalnumberoffourandseven(x * 10 + 7,n,a)
    return a
#----------------------
def luckynum(x,l,r,a):
    if x>=l and x<=r:
        a.append(x)
    if x>r:
        a.append(x)
        return a
    if x < 1e10:
            luckynum(x * 10 + 4, l,r,a)
            luckynum(x * 10 + 7, l,r,a)
    return a

def luckynuber(x, n, a):
    p = set(str(x))
    if len(p) <= 2:
        a.append(x)
    if x < n:
        luckynuber(x + 1, n, a)
    return a


# ------------------------------------------------------interactive problems

def interact(type, x):
    if type == "r":
        inp = input()
        return inp.strip()
    else:
        print(x, flush=True)


# ------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
    # Initialize result
    count = 0

    # Keep dividing n by
    # 5 & update Count
    while (n >= 5):
        n //= 5
        count += n

    return count


# -----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
    if len(arr) > 1:

        # Finding the mid of the array
        mid = len(arr) // 2

        # Dividing the array elements
        L = arr[:mid]

        # into 2 halves
        R = arr[mid:]

        # Sorting the first half
        mergeSort(L)

        # Sorting the second half
        mergeSort(R)

        i = j = k = 0

        # Copy data to temp arrays L[] and R[]
        while i < len(L) and j < len(R):
            if L[i] < R[j]:
                arr[k] = L[i]
                i += 1
            else:
                arr[k] = R[j]
                j += 1
            k += 1

        # Checking if any element was left
        while i < len(L):
            arr[k] = L[i]
            i += 1
            k += 1

        while j < len(R):
            arr[k] = R[j]
            j += 1
            k += 1


# -----------------------------------------------lucky number with two lucky any digits
res = set()


def solven(p, l, a, b, n):  # given number
    if p > n or l > 10:
        return
    if p > 0:
        res.add(p)
    solven(p * 10 + a, l + 1, a, b, n)
    solven(p * 10 + b, l + 1, a, b, n)


# problem
"""
n = int(input())
for a in range(0, 10):
    for b in range(0, a):
        solve(0, 0)
print(len(res))
"""


# Python3 program to find all subsets
# by backtracking.

# In the array A at every step we have two
# choices for each element either we can
# ignore the element or we can include the
# element in our subset
def subsetsUtil(A, subset, index, d):
    print(*subset)
    s = sum(subset)
    d.append(s)
    for i in range(index, len(A)):
        # include the A[i] in subset.
        subset.append(A[i])

        # move onto the next element.
        subsetsUtil(A, subset, i + 1, d)

        # exclude the A[i] from subset and
        # triggers backtracking.
        subset.pop(-1)
    return d


def subsetSums(arr, l, r, d, sum=0):
    if l > r:
        d.append(sum)
        return
    subsetSums(arr, l + 1, r, d, sum + arr[l])

    # Subset excluding arr[l]
    subsetSums(arr, l + 1, r, d, sum)
    return d


def print_factors(x):
    factors = []
    for i in range(1, x + 1):
        if x % i == 0:
            factors.append(i)
    return (factors)


# -----------------------------------------------
def calc(X, d, ans, D):
    # print(X,d)
    if len(X) == 0:
        return

    i = X.index(max(X))
    ans[D[max(X)]] = d

    Y = X[:i]
    Z = X[i + 1:]

    calc(Y, d + 1, ans, D)
    calc(Z, d + 1, ans, D)


# ---------------------------------------

def factorization(n, l):
    c = n
    if prime(n) == True:
        l.append(n)
        return l
    for i in range(2, c):
        if n == 1:
            break
        while n % i == 0:
            l.append(i)
            n = n // i
    return l


# endregion------------------------------


def good(b):
    l = []
    i = 0
    while (len(b) != 0):
        if b[i] < b[len(b) - 1 - i]:
            l.append(b[i])
            b.remove(b[i])
        else:
            l.append(b[len(b) - 1 - i])
            b.remove(b[len(b) - 1 - i])
    if l == sorted(l):
        # print(l)
        return True
    return False


# arr=[]
# print(good(arr))
def generate(st, s):
    if len(s) == 0:
        return

    # If current string is not already present.
    if s not in st:
        st.add(s)

        # Traverse current string, one by one
        # remove every character and recur.
        for i in range(len(s)):
            t = list(s).copy()
            t.remove(s[i])
            t = ''.join(t)
            generate(st, t)
    return
#=--------------------------------------------longest increasing subsequence

def largestincreasingsubsequence(A):
    l = [1]*len(A)
    sub=[]
    for i in range(1,len(l)):
        for k in range(i):
            if A[k]<A[i]:
                sub.append(l[k])
        l[i]=1+max(sub,default=0)
    return max(l,default=0)

#----------------------------------longest palindromic substring




# Python3 program for the
# above approach

# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums, N):
    # Stores the prefix
    # sum of nums[]
    prefix_sum = 0

    # Stores the bitwise OR of
    # sum of each subsequence
    result = 0

    # Iterate through array nums[]
    for i in range(N):
        # Bits set in nums[i] are
        # also set in result
        result |= nums[i]

        # Calculate prefix_sum
        prefix_sum += nums[i]

        # Bits set in prefix_sum
        # are also set in result
        result |= prefix_sum

    # Return the result
    return result

#l=[]
def OR(a, n):
    ans = a[0]
    for i in range(1, n):
        ans |= a[i]
        #l.append(ans)
    return ans
#print(prime(12345678987766))



def toString(List):
    return ''.join(List)


# Function to print permutations of string
# This function takes three parameters:
# 1. String
# 2. Starting index of the string
# 3. Ending index of the string.
p=[]
def permute(a, l, r):
    if l == r:
        p.append(toString(a))
    else:
        for i in range(l, r + 1):
            a[l], a[i] = a[i], a[l]
            permute(a, l + 1, r)
            a[l], a[i] = a[i], a[l]  # backtrack

# Function to find square root of
# given number upto given precision
def squareRoot(number, precision):
    start = 0
    end, ans = number, 1

    # For computing integral part
    # of square root of number
    while (start <= end):
        mid = int((start + end) / 2)

        if (mid * mid == number):
            ans = mid
            break

        # incrementing start if integral
        # part lies on right side of the mid
        if (mid * mid < number):
            start = mid + 1


        # decrementing end if integral part
        # lies on the left side of the mid
        else:
            end = mid - 1

    # For computing the fractional part
    # of square root upto given precision
    increment = 0.1
    for i in range(0, precision):
        while (ans * ans <= number):
            ans += increment

        # loop terminates when ans * ans > number
        ans = ans - increment
        increment = increment / 10

    return ans
#for i in range(200):
#    if sumofdigits(i)==10:
        #print(i)

import math


def countRectangles(l, w):
    # if we take gcd(l, w), this
    # will be largest possible
    # side for suare, hence minimum
    # number of square.
    squareSide = math.gcd(l, w)

    # Number of squares.
    return int((l * w) / (squareSide * squareSide))


# Function that count the
# total numbersProgram between L
# and R which have all the
# digit same
def count_same_digit(L, R):
    tmp = 0
    ans = 0

    # length of R
    n = int(math.log10(R) + 1)

    for i in range(0, n):

        # tmp has all digits as 1
        tmp = tmp * 10 + 1

        # For each multiple
        # of tmp in range 1 to 9,
        # check if it present
        # in range [L, R]
        for j in range(1, 10):

            if (L <= (tmp * j) and (tmp * j) <= R):
                #print(tmp*j)
                # Increment the required count
                ans += 1

    return ans
#l=[]
#for i in range(2,31):
#    l.append(2**i)
def main():
    for _ in range(inpu()):
        n=inpu()
        a,b=inp(),inp()
        l = []
        for i in range(n):
            if a[i] != b[i]:
                l += [i + 1, 1, i + 1]
        print(len(l),*l)
if __name__ == '__main__':
    main()