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.
def solve(a,b,n,ans):
    orig = a[:]
    moves = []
    end = n-1
    start = 0
    visited = set()
    end1 = n-1
    done = 0
    while end1 >= 0:
        if done%2 == 0:
            start_val = a[start]
            end_val = a[end]
        else:
            start_val = str(int(not int(a[start])))
            end_val = str(int(not int(a[end])))
            
        if end_val != b[end1]:

            #print(start_val,end_val,start,end,'here')
            if start_val == end_val:
                moves.append(abs(end-start)+1)
                done += 1
                visited.add(end)
                end,start = start,end
                end1 -= 1
                if end < start:
                    end += 1
                else:
                    end -= 1
            else:
                moves.append(1)
                a[start] = str(int(not int(a[start])))

            #print('after',a)
        else:
            visited.add(end)
            if end < start:
                end += 1
            else:
                end -= 1

            end1 -= 1

    ans.append(moves)

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

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

        print()
        


main()
