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 math as mt
def tran(s1,s2,n):
    an=[]
    for i in range(n):
        if s1[n-i-1]==s2[n-i-1]:
            pass
        else:
            if s2[n-i-1]!=s1[0]:
                an.append(n-i)
                s11=str(s1)
                ss=s1[:n-i]
                Sar=[]
                for j in ss:
                    if j=='1':
                        Sar.append('0')
                    else:
                        Sar.append('1')
                Sar.reverse()
                s1=''
                for k in Sar:
                    s1=s1+k
                s1=s1+s11[n-i:]
            else:
                an.append(1)
                s11=str(s1)
                if s1[0]=='0':
                    s1=s1.replace('0','1',1)
                else:
                    s1=s1.replace('1','0',1)
                an.append(n-i)
                ss = s1[:n - i]
                Sar = []
                for j in ss:
                    if j == '1':
                        Sar.append('0')
                    else:
                        Sar.append('1')
                Sar.reverse()
                s1 = ''
                for k in Sar:
                    s1 = s1 + k
                s1 = s1 + s11[n - i:]

    if len(an)==0:
        return []
    else:
        return an

if __name__ == '__main__':
    t = int(input())
    ans = []
    for i in range(t):
        n= int(input())
        s1=input()
        s2=input()
        r = tran(s1,s2,n)
        ans.append(r)
    for i in ans:
        if len(i)==0:
            print(0)
        else:
            p=len(i)
            i.insert(0,p)
            print(*i)