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 sys
import heapq, functools, collections
import math, random
from collections import Counter, defaultdict

# available on Google, not available on Codeforces
# import numpy as np
# import scipy


def solve(arr, brr):  # fix inputs here
    console("----- solving ------")
    arr = [int(x) for x in arr]
    brr = [int(x) for x in brr]
    console(brr, "**")
    console(arr)

    ptr = len(arr) - 1

    res = []
    # cnt = 0
    if arr == brr:
        return "0"

    def operate(crr):
        return [x^1 for x in crr][::-1]

    while ptr >= 0:
        if arr[ptr] == brr[ptr]:
            ptr -= 1
            continue
        if arr[0] == brr[ptr]:
            arr[0] = arr[0]^1
            res.append(1)
            console(arr, 1)

        res.append(ptr+1)
        arr[:ptr+1] = operate(arr[:ptr+1])
        console(arr, ptr+1)

    return str(len(res)) + " " + " ".join([str(x) for x in res])


def console(*args):  # the judge will not read these print statement
    # print('\033[36m', *args, '\033[0m', file=sys.stderr)
    return

# fast read all
# sys.stdin.readlines()

for case_num in range(int(input())):
    # read line as a string
    # strr = input()

    # read line as an integer
    _ = int(input())
    
    # read one line and parse each word as a string
    arr = input()
    brr = input()

    # read one line and parse each word as an integer
    # lst = list(map(int,input().split()))

    # read matrix and parse as integers (after reading read nrows)
    # lst = list(map(int,input().split()))
    # nrows = lst[0]  # index containing information, please change
    # grid = []
    # for _ in range(nrows):
    #     grid.append(list(map(int,input().split())))

    res = solve(arr, brr)  # please change
    
    # Google - case number required
    # print("Case #{}: {}".format(case_num+1, res))

    # Codeforces - no case number required
    print(res)
