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.
from collections import *
import sys
try: inp = raw_input
except: inp = input
def err(s):
    sys.stderr.write('{}\n'.format(s))

def ni():
    return int(inp())

def nl():
    return [int(_) for _ in inp().split()]

T = ni()
def swap(ch):
    return '1' if ch == '0' else '0'

def get_a_l(a, flips):
    if flips%2 == 0:
        return a[0]
    else:
        return swap(a[-1])

def get_a_r(a, flips):
    if flips%2 == 0:
        return a[-1]
    else:
        return swap(a[0])

def solve(a, b):
    a = deque(a)
    out = []
    flips = 0
    for i in range(len(a) - 1, -1, -1):
        if (get_a_r(a, flips) == b[i]):
            if flips%2 == 0:
                a.pop()
            else:
                a.popleft()
            continue
        if (get_a_l(a, flips) != b[i]):
            out.append(i+1)
        else:
            out.append(1)
            out.append(i+1)
        flips += 1
        if flips%2 == 0:
            a.pop()
        else:
            a.popleft()
    return out

for _ in range(T):
    N = ni()
    a = list(inp())
    b = list(inp())
    out = solve(a, b)
    print(' '.join(map(str, [len(out)] + out)))
    
    
