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 _ in range(t):
    n=int(input())
    a=input()
    b=input()
    
    aToAllOnesOrZeros=[]
    for i in range(n-1):
        if a[i]!=a[i+1]: #flip up to a[i]. Then a[0,1,...i+1] will be uniform. a will become all = to a[-1]
            aToAllOnesOrZeros.append(i+1)
    bToAllOnesOrZeros=[]
    for i in range(n-1):
        if b[i]!=b[i+1]: #b will become all = to b[-1]
            bToAllOnesOrZeros.append(i+1)
    res=aToAllOnesOrZeros
    if a[-1]!=b[-1]: #flip all aTransformed so that aTransformed==bTransformed 
        res.append(n)
    for i in range(len(bToAllOnesOrZeros)-1,-1,-1):
        res.append(bToAllOnesOrZeros[i]) #reverse the process
    l=len(res)
    res.insert(0,l)
    print(' '.join([str(x) for x in res]))






#def flip(s,l):
#    S=list(s[:l])
#    for i in range(l):
#        S[i]='0' if S[i]=='1' else '1'
#    return ''.join(S)+s[l:]
#
#y=a
#for i in range(n-1):
#    if y[i]!=y[i+1]:
#        y=flip(y,i+1)
#print(y) #y will be all a[-1]
#
#z=b
#for i in range(n-1):
#    if z[i]!=z[i+1]:
#        z=flip(z,i+1)
#print(z) #z will be all b[-1]