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 string
import math
from collections import defaultdict
from functools import lru_cache
from collections import Counter
from fractions import Fraction

def mi(s):
    return map(int, s.strip().split())

def lmi(s):
    return list(mi(s))

def tmi(s):
    return tuple(mi(s))

def mf(f, s):
    return map(f, s)

def lmf(f, s):
    return list(mf(f, s))

def js(lst):
    return " ".join(str(d) for d in lst)

def jsns(lst):
    return "".join(str(d) for d in lst)

def line():
    return sys.stdin.readline().strip()

def linesp():
    return line().split()

def iline():
    return int(line())

def mat(n):
    matr = []
    for _ in range(n):
        matr.append(linesp())
    return matr

def matns(n):
    mat = []
    for _ in range(n):
        mat.append([c for c in line()])
    return mat

def mati(n):
    mat = []
    for _ in range(n):
       mat.append(lmi(line())) 
    return mat

def pmat(mat):
    for row in mat:
        print(js(row))

def flip(arr, i, j):
    while i < j + 1:
        arr[i] = 1 - arr[i]
        i += 1

def reverse(arr, i, j):
    while i < j:
        arr[i], arr[j] = arr[j], arr[i]
        i += 1
        j -= 1

def solve(A, B):
    A = [int(c) for c in A]
    B = [int(c) for c in B]

    moves = []
    for i in range(len(A) - 1, -1, -1):
        if i == 0:
            if A[i] == B[i]:
                break
            else:
                flip(A, 0, 0)
                moves.append(1)
        elif A[i] == B[i]:
            continue
        else:
            # We want to get B[i].
            want = B[i]
            if A[0] == B[i]:
                moves.append(1)
            moves.append(i + 1)

            flip(A, 0, i)
            reverse(A, 0, i)
    print(len(moves), js(moves))

def main():
    t = iline()
    for _ in range(t):
        line()
        solve(line(), line())
main()
