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:
T = int(input())
for t in range(T):
    n = int(input())
    a = input()
    b = input()
    isInv = False
    l=0
    r= n-1
    i = n-1
    ans = 0
    res = []
    while(i>=0):
        if(isInv):
            if(b[i]==a[l]):
                if(b[i] != a[r]):
                    res.append(1)
                    ans+=1
                ans+=1
                res.append(r-l+1)
                r-=1
                isInv = not(isInv)
            else:
                l+=1

        else:
            if(b[i] != a[r]):
                if(b[i]==a[l]):
                    res.append(1)
                    ans+=1
                ans+=1
                res.append(r-l+1)
                l+=1
                isInv = not(isInv)
            else:
                r-=1
        i-=1
    print(ans, end=" ")
    for i in range(ans):
        print(res[i], end=" ")
    print()