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 PYTHON):

Solution:
from __future__ import division, print_function


def main():
    # Template 1.0
    import sys, re, math
    from collections import deque, defaultdict, Counter, OrderedDict
    from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, floor
    from heapq import heappush, heappop, heapify, nlargest, nsmallest
    from itertools import permutations
    def STR():
        return list(input())

    def INT():
        return int(input())

    def MAP():
        return map(int, input().split())

    def LIST():
        return list(map(int, input().split()))

    def list2d(a, b, c):
        return [[c] * b for i in range(a)]

    def sortListWithIndex(listOfTuples, idx):
        return (sorted(listOfTuples, key=lambda x: x[idx]))

    def sortDictWithVal(passedDic):
        temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))
        toret = {}
        for tup in temp:
            toret[tup[0]] = tup[1]
        return toret

    def sortDictWithKey(passedDic):
        return dict(OrderedDict(sorted(passedDic.items())))

    INF = float('inf')
    mod = 10 ** 9 + 7

    t = INT()

    while(t!=0):
        n =INT()

        a = list(map(int, list(input())))
        b = list(map(int, list(input())))

        ans = []

        for i in range(n-1, -1, -1):
            if(a[i]!=b[i]):
                if(a[0]==b[i]):
                    ans.append(1)
                    ans.append(i+1)
                    #change
                    a[0]^=1
                    temp = a[:i+1]

                    for j in range(len(temp)):
                        temp[j]^=1
                    temp = temp[::-1]
                    for j in range(len(temp)):
                        a[j] = temp[j]
                else:
                    ans.append(i+1)
                    temp = a[:i + 1]

                    for j in range(len(temp)):
                        temp[j] ^= 1
                    temp = temp[::-1]
                    for j in range(len(temp)):
                        a[j] = temp[j]
        print(len(ans), *ans)

        t-=1

######## Python 2 and 3 footer by Pajenegod and c1729
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# Cout implemented in Python
import sys


class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self


cout = ostream()
endl = '\n'


# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A


if __name__ == "__main__":
    main()