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:
"Codeforces Round #658 (Div. 2)"
"A. Common Subsequence"
# y=int(input())
# for i in range(y):
#     n,m=map(int,input().split())
#     a=map(int,input().split())
#     b=map(int,input().split())
#     sa=set(a)
#     sb=set(b)
#     sc=set()
#     sc=sa.intersection(sb)
#     if len(sc)==0:
#         print("NO")
#     else:
#         print("YES")
#         print(1," ",sep="",end="")
#         for j in sc:
#             print(j)
#             break    
"B. Sequential Nim"
# y=int(input())
# for i in range(y):
#     n=int(input())
#     a=list(map(int,input().split()))
#     if sum(a)==n:
#         if n%2==0:
#             print("Second")
#         else:
#             print("First")
#     else:
#         count_ones=0
#         for j in range(n):
#             if a[j]>1:
#                 break
#             else:
#                 count_ones+=1
#         if count_ones%2==0:
#             print("First")
#         else:
#             print("Second")      
"C1. Prefix Flip (Easy Version)"
def func(a,n):
    for i in range(n):
        a[i]=1-a[i]
    n1=int(n/2)    
    for i in range(n1):
        t=a[i]
        a[i]=a[n-1-i]
        a[n-1-i]=t    

y=int(input())
for i in range(y):
    n=int(input())
    a=list(map(int,list(input())))
    b=list(map(int,list(input())))
    ans=[]
    for j in range(1,n):
        # print(j,a[-j],b[-j],a[0])
        if a[-j]!=b[-j]:
            if a[0]!=b[-j]:
                func(a,n-j+1)
                ans.append(n-j+1)
            else:
                func(a,1)
                ans.append(1)
                func(a,n-j+1)
                ans.append(n-j+1)
    if a[0]!=b[0]:
        ans.append(1)    
    print(len(ans)," ",end="",sep="")
    for j in ans:
        print(j," ",sep="",end="")
    print()