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:

from __future__ import division
import sys
input = sys.stdin.readline
import math
from math import sqrt, floor, ceil
from collections import Counter



############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
def insr2():
    s = input()
    return(s.split(" "))

def prime_factorization(n):

    if n == 1:
        return [1]

    ans=[]
    i = 2
    cap = sqrt(n)
    while i <= cap:
        if n % i == 0:
            ans.append(i)
            n = n//i
            cap=sqrt(n)
        else:
            i += 1
    if n > 1:
        ans.append(n)
    return ans

def binomial(n, k):
    if n == 1 or n == k:
        return 1

    if k > n:
        return 0       
    else:
        a = math.factorial(n)
        b = math.factorial(k)
        c = math.factorial(n-k)
        div = a // (b * c)
        return div 

for _ in range(inp()):
    n= inp()
    a = insr()
    b = insr()

    opsa, opsb=[],[]

    for i in range(n-1):
        if a[i] != a[i+1]:
            opsa.append(i+1)
        if b[i] != b[i+1]:
            opsb.append(i+1)

    if b[-1] != a[-1]:
        opsb.append(n)

    ans = opsa + opsb[::-1]
    ans.insert(0, len(ans))
    
    print(' '.join([str(a) for a in ans]))