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:
import sys
r=sys.stdin.readline
def revAndnot(bit,idx):
    tmp=""
    for i in range(idx):
        tmp+='1' if bit[i]=='0' else '0'
    return tmp[::-1]+bit[idx:]
for _ in range(int(r())):
    n=int(r())  #length of binary
    a=r().strip()
    b=r().strip()
    ans=[]
    """ 3n operation method
    for i in range(n):
        if a[i]!=b[i]:
            if i>0:
                ans.append(str(i+1))
            ans.append('1')
            if i>0:
                ans.append(str(i+1))
    """
    """2n operation method
    for i in range(n-1,-1,-1):
        if b[i]==a[i]:
            continue
        elif a[0]==b[i]:#2op
            a=revAndnot(a,1)
            a=revAndnot(a,i+1)
            ans.append(str(1))
            ans.append(str(i+1))
        else:
            a=revAndnot(a,i+1)
            ans.append(str(i+1))
    """
    flip=False
    idx=0
    for i in range(n-1,-1,-1):
        if flip^(a[idx]==b[i]):
            ans.append(str(1))
        ans.append(str(i+1))
        if flip: idx-=i
        else: idx+=i
        flip^=True

    print(len(ans),end=" ")
    print(' '.join(ans))