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:
from collections import defaultdict as dd,deque as dq
import sys
input=sys.stdin.readline
t=int(input())
while t:
    n=int(input())
    a=list(map(int,input().strip()))
    b=list(map(int,input().strip()))
    q=dq(a)
    rev=0
    par=0
    last=n-1
    i=0
    j=n-1
    ans=[]
    while q:
        if(rev==0):
            if(q):
                q[0]=q[0]^par
                lo=0
                if(q[0]==b[last]):
                    ans.append(1)
                    
                    lo=1
                if(q):
                    rev=1-rev
                    par=1-par
                    ans.append(len(q))
                q.popleft()
                last-=1
        else:
            j=len(q)-1
            if(q):
                q[j]=q[j]^par
                lo=0
                if(q[j]==b[last]):
                    j-=1
                    ans.append(1)
                    lo=1
                if(q):
                    rev=1-rev
                    par=1-par
                    ans.append(len(q))
                q.pop()
                last-=1
    print(len(ans),*ans)
    t-=1