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 copy import copy

def main():
    t = int(input())
    for _ in range(t):
        ans = []
        n = int(input())
        a = [int(i) for i in input()]
        b = [int(i) for i in input()]

        def change(c):
            if c == 1:
                return 0
            else:
                return 1

        # def whole_flip(i):
        #     n = i+1
        #     ans.append(1)
        #     ans.append(n)
        #     a[0] = change(a[0])
        #     for j in range(n//2):
        #         a[j], a[n-j-1] = a[n-j-1], a[j]
        #         a[j] = change(a[j])
        #         a[n-j-1] = change(a[n-j-1])
        #     if n % 2 != 0:
        #         a[(n//2) + 1] = change(a[(n//2) + 1])
        
        def flip(a, i):
            n = i+1
            ans.append(n)
            temp = copy(a)
            for i in range(n):
                a[i] = temp[n-i-1]
                a[i] = change(a[i])
            # print(a)

        
        for i in range(n-1, 0, -1):
            if a[i] != b[i]:
                start = a[0]
                if start != a[i]:
                    ans.append(1)
                    a[0] = change(a[0])
                    flip(a, i)
                else:
                    flip(a, i)
            # print(i)
            # print(a)
        
        if a[0] != b[0]:
            ans.append(1)

        print(len(ans), end=' ')
        for i in ans:
            print(i, end=' ')
        print()

main()