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 collections import defaultdict as dd
from collections import deque
import bisect
import heapq

def ri():
    return int(input())

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


def solve():
    n = ri()
    s = list(map(int, input()))
    t = list(map(int, input()))
    moves = []

    prev_flip = n - 1
    rng = [0, n - 1]

    def UPDATE(pos, prev_flip, rng):
        d = prev_flip - pos
        prev_flip = pos
        a, b = rng
        if a < b:
            rng = [b - d, a]
        else:
            rng = [b + d, a]
        return prev_flip, rng

    def STATUS(pos, rng):
        a, b = rng
        if a <= b:
            orig = s[a + pos]
            return orig
        else:
            orig = s[a - pos]
            return 1 - orig # odd number of flips

    for pos in range(n - 1, 0, -1):
        if STATUS(pos, rng) == t[pos]:
            continue
        elif STATUS(0, rng) != t[pos]:
            moves.append(pos + 1)
            prev_flip, rng = UPDATE(pos, prev_flip, rng)
        else:
            moves.append(1)
            moves.append(pos + 1)
            prev_flip, rng = UPDATE(pos, prev_flip, rng)

    if STATUS(0, rng) != t[0]:
        moves.append(1)

    print (len(moves), *moves)



mode = 'T'

if mode == 'T':
    t = ri()
    for i in range(t):
        solve()
else:
    solve()