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 os,io
from sys import stdout
import collections
# import random
# import math
# from operator import itemgetter
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# from decimal import Decimal
# import heapq
# from functools import lru_cache

# import sys
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(102400000)

# from functools import lru_cache
# @lru_cache(maxsize=None)

######################
# --- Maths Fns  --- #
######################
def primes(n):
    sieve = [True] * n
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
    return [2] + [i for i in range(3,n,2) if sieve[i]]

def binomial_coefficient(n, k):
    if 0 <= k <= n:
        ntok = 1
        ktok = 1
        for t in range(1, min(k, n - k) + 1):
            ntok *= n
            ktok *= t
            n -= 1
        return ntok // ktok
    else:
        return 0

def powerOfK(k, max):
    if k == 1:
        return [1]
    if k == -1:
        return [-1, 1]

    result = []
    n = 1
    while n <= max:
        result.append(n)
        n *= k
    return result

def divisors(n):
    i = 1
    result = []
    while i*i <= n:
        if n%i == 0:
            if n/i == i:
                result.append(i)
            else:
                result.append(i)
                result.append(n/i)
        i+=1
    return result

def distance(xa, ya, xb, yb):
    return ((ya-yb)**2 + (xa-xb)**2)**(0.5)

# @lru_cache(maxsize=None)
def digitsSum(n):
    if n == 0:
        return 0
    r = 0
    while n > 0:
        r += n % 10
        n //= 10
    return r

######################
# ---- GRID Fns ---- #
######################
def isValid(i, j, n, m):
    return i >= 0 and i < n and j >= 0 and j < m

def print_grid(grid):
    for line in grid:
        print(" ".join(map(str,line)))

######################
# ---- MISC Fns ---- #
######################
def kadane(a,size):
    max_so_far = 0
    max_ending_here = 0

    for i in range(0, size):
        max_ending_here = max_ending_here + a[i]
        if (max_so_far < max_ending_here):
            max_so_far = max_ending_here

        if max_ending_here < 0:
            max_ending_here = 0
    return max_so_far

def prefixSum(arr):
    for i in range(1, len(arr)):
        arr[i] = arr[i] + arr[i-1]
    return arr

def ceil(n, d):
    if n % d == 0:
        return n // d
    else:
        return (n // d) + 1

def _abs(a, b):
    return int(abs(a - b))

def evenodd(l):
    even = [e for e in l if e & 1 == 0]
    odd = [e for e in l if e & 1]
    return (even, odd)

# INPUTS  --------------------------
# s = input().decode('utf-8').strip()
# n = int(input())
# l = list(map(int, input().split()))
# t = int(input())
# for _ in range(t):
# for _ in range(int(input())):

def op(arr):
    for i in range(len(arr)):
        if arr[i] == 1:
            arr[i] = 0
        else:
            arr[i] = 1
    r = arr[::-1]
    return r

for _ in range(int(input())):
    n = int(input())
    a = list(map(int, list(input().decode('utf-8').strip())))
    b = list(map(int, list(input().decode('utf-8').strip())))

    fr = []
    for e in b[::-1]:
        if e == 1:
            fr.append(0)
        else:
            fr.append(1)

    i = len(a) - 1
    operations = []
    while i >= 0:
        if a[i] == b[i]:
            i -= 1
            continue

        if a[0] != a[i]:
            if a[0] == 1:
                a[0] = 0
            else:
                a[0] = 1
            operations.append(1)

        na = op(a[:i+1]) + a[i+1:]
        operations.append(i+1)
        i -= 1
        a = na


    print(len(operations), " ".join(list(map(str, operations))))
