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:
class Solution:
    def __init__(self, debug=False, n=None, a=None, b=None):
        if n is None:
            n, a, b = Solution.input()
        self.n = n
        self.a = a
        self.b = b

        self.real = [x for x in self.a]
        self.inverse = [not x for x in self.a]
        self.curr_zero = 0
        self.curr_last = self.n - 1
        self.real_idx = self.n - 1
        self.inv = False

        self.debug = debug
        if debug:
            self.current_state = [x for x in self.a]

        self.operations = [self.n + 1000]

    @staticmethod
    def input():
        n = int(input())
        a = [x == '1' for x in input()]
        b = [x == '1' for x in input()]
        return n, a, b

    def iteration(self):
        if self.inv:
            if self.inverse[self.curr_last] != self.b[self.real_idx]:
                if self.inverse[self.curr_zero] != self.inverse[self.curr_last]:
                    self.operation(0)
                    self.real[self.curr_zero], self.inverse[self.curr_zero] = self.inverse[self.curr_zero], self.real[
                        self.curr_zero]
                self.operation(self.real_idx)
                self.inv = False
                self.curr_zero, self.curr_last = self.curr_last, self.curr_zero
        else:
            if self.real[self.curr_last] != self.b[self.real_idx]:
                if self.real[self.curr_zero] != self.real[self.curr_last]:
                    self.operation(0)
                    self.real[self.curr_zero], self.inverse[self.curr_zero] = self.inverse[self.curr_zero], self.real[
                        self.curr_zero]
                self.operation(self.real_idx)
                self.inv = True
                self.curr_zero, self.curr_last = self.curr_last, self.curr_zero

        if self.inv:
            self.curr_last += 1
        else:
            self.curr_last -= 1

        self.real_idx -= 1

    def solve(self):
        while self.real_idx >= 0:
            self.iteration()
        return self.return_answer()

    def operation(self, i):
        self.operations.append(i + 1)
        if self.debug:
            self.current_state[:i + 1] = [not x for x in self.current_state[:i + 1]][::-1]

    def return_answer(self):
        ans = f'{len(self.operations) - 1} {" ".join(str(x) for x in self.operations[1:])}'
        print(ans)
        return ans


def solve():
    n = int(input())
    a = [int(x) for x in input()]
    b = [int(x) for x in input()]

    # state
    real = [x for x in a]
    inverse = [(x + 1) % 2 for x in a]
    inv_idx = 0
    real_idx = n - 1
    inv = False

    operations = []
    # fix all
    seq = []
    while real_idx > 0:
        if not inv:
            if real[real_idx] == b[real_idx]:
                real_idx -= 1
                inv_idx += 1
                seq.append(real[real_idx])
                continue

            if real[0] != real[real_idx]:
                operations.append(1)
                real[0], inverse[-1] = inverse[-1], real[0]

            operations.append(real_idx + 1)
            real[real_idx] = b[real_idx]
            seq.append(real[real_idx])
            inverse[inv_idx] = (real[real_idx] + 1) % 2
            inv = True
        else:
            if inverse[inv_idx] == b[real_idx]:
                real_idx -= 1
                inv_idx += 1
                seq.append(inverse[inv_idx])
                continue
            if inverse[-1] != inverse[inv_idx]:
                operations.append(1)
                real[0], inverse[-1] = inverse[-1], real[0]

            operations.append(real_idx + 1)
            inverse[inv_idx] = b[real_idx]
            real[real_idx] = (b[real_idx] + 1) % 2
            seq.append(inverse[inv_idx])
            inv = False
        real_idx -= 1
        inv_idx += 1

    # fix 0
    if (inv and inverse[-1] != b[0]) or (real[0] != b[0] and not inv):
        operations.append(1)
    seq.append(b[0])
    print(seq, b)
    print(len(operations), *operations)


if __name__ == '__main__':
    for _ in range(int(input())):
        n, a, b = Solution.input()
        Solution(False, n, a, b).solve()