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.
t = int(input().split()[0])
for case in range(t):
    n = int(input().split()[0])
    a = [int(x) for x in input().split()[0]]
    b = [int(x) for x in input().split()[0]]
    l = 0
    r = n-1
    last = n-1
    while r > -1:
        if a[r] == b[r]:
            r -= 1
            last -= 1
        else:
            break
    ans = []
    flips = 0    
    ltor = True
    while l-1 < r:
        if ltor:
            if (a[l] + flips) % 2 == b[last]:
                ans.append(1)
                ans.append(last+1)
            else:
                ans.append(last+1)
            flips += 1
            l += 1
            last -= 1
            while l < n and (a[l] + flips) % 2 == b[last]:
                l += 1
                last -= 1               
        else:
            if (a[r] + flips) % 2 == b[last]:
                ans.append(1)
                ans.append(last+1)
            else:
                ans.append(last+1)
            flips += 1
            r -= 1
            last -= 1
            while r > -1 and (a[r] + flips) % 2 == b[last]:
                r -= 1
                last -= 1
        ltor = not ltor
    print(str(len(ans)) + " ",end="")
    print(*ans)
