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.
# cook your dish here
#code
#    ___________________________________
#   |                                   |
#   |                                   | 
#   |        _,     _   _     ,_        |
#   |    .-'` /     \'-'/     \ `'-.    |
#   |   /    |      |   |      |    \   |
#   |  ;      \_  _/     \_  _/      ;  |
#   | |         ``         ``         | |
#   | |                               | |
#   |  ;    .-.   .-.   .-.   .-.    ;  |
#   |   \  (   '.'   \ /   '.'   )  /   |
#   |    '-.;         V         ;.-'    |
#   |        `                 `        |
#   |                                   |
#   |___________________________________|
#   |                                   |
#   |  Author      :   Ramzz            |
#   |  Created On  :   21-07-2020       |
#   |___________________________________|
#
#    _ __ __ _ _ __ ___  ________
#   | '__/ _` | '_ ` _ \|_  /_  /
#   | | | (_| | | | | | |/ / / / 
#   |_|  \__,_|_| |_| |_/___/___|
#

import math
import collections
from sys import stdin,stdout,setrecursionlimit
from bisect import bisect_left as bsl
from bisect import bisect_right as bsr
import heapq as hq
setrecursionlimit(2**20)

t = 1
t = int(stdin.readline())

for _ in range(t):
    n = int(stdin.readline())
    a = str(stdin.readline().strip('\n'))
    b = str(stdin.readline().strip('\n'))
    #a = list(map(int, stdin.readline().rstrip().split()))
    
    ans = []
    cur = a[0]
    l = []
    inv = 0
    in0 = 0
    for i in range(n-1,-1,-1):
        if(b[i]==a[0]):
            ans.append(1)
            ans.append(i+1)
            for i in [ans[-2],ans[-1]]:
                tmp = ''
                for j in range(i):
                    if(a[j]=='1'):
                        tmp+='0'
                    else:
                        tmp+='1'
                tmp = tmp[::-1]
                a = tmp+a[i:]
        else:
            ans.append(i+1)
            for i in [ans[-1]]:
                tmp = ''
                for j in range(i):
                    if(a[j]=='1'):
                        tmp+='0'
                    else:
                        tmp+='1'
                tmp = tmp[::-1]
                a = tmp+a[i:]
                
    #print(a)
    print(len(ans),end=' ')
    for i in ans:
        print(i,end=' ')
    print()
    #print(ans)