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
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__

def gift():
    for _ in range(t):
        n=int(input())

        s1=list(input())
        s2=list(input())
        
        res=[]
        last = s1[0]
        for i in range(1, n):
            #print(a)
            if s1[i] != last:
                res.append(i)
                last = s1[i]
        for i in range(n - 1, -1, -1):
            #print(a)
            if s2[i] != last:
                res.append(i + 1)
                last = s2[i]        
     


        yield " ".join(str(x) for x in [len(res)]+res)
 


if __name__ == '__main__':
    t= int(input())
    ans = gift()
    print(*ans,sep='\n')
            


#"{} {} {}".format(maxele,minele,minele)
# 001011

# 111011
# 111011

# 000100
# 001000

# 101011

# 010100
# 001010



#101111
#010000
#000010


#110100
#001011

#011
#10100
#11100


#11111
#00011

#11011
#00000