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:
for _ in range(int(input())):
    N=int(input())
    A=list(input())
    B=list(input())
    countlist=[]
    countlistB=[]
    count=1
    for i in range(1,N):
        if(A[i]!=A[i-1]):
            countlist.append(count)
            count=1
        else:
            count+=1
    countlist.append(count)
    countB=1
    for i in range(1,N):
        if(B[i]!=B[i-1]):
            countlistB.append(countB)
            countB=1
        else:
            countB+=1
    countlistB.append(countB)
    # print(countlist)
    prefix=[]
    prefix.append(countlist[0])
    for i in range(1,len(countlist)):
        prefix.append(countlist[i]+prefix[-1])
    # print(prefix)
    prefixB=[]
    prefixB.append(countlistB[0])
    for i in range(1,len(countlistB)):
        prefixB.append(countlistB[i]+prefixB[-1])
    # print(prefixB)
    ans=0
    if(len(countlist)%2==0):
        if(A[0]==B[-1]):
            ans=len(prefix)+len(prefixB)-1
            print(ans,end=" ")
            for i in range(len(prefix)):
                print(prefix[i],end=" ")
            for i in range(len(prefixB)-2,-1,-1):
                print(prefixB[i],end=" ")
            print()
        else:
            ans=len(prefix)+len(prefixB)
            print(ans,end=" ")
            for i in range(len(prefix)):
                print(prefix[i],end=" ")
            for i in range(len(prefixB)-1,-1,-1):
                print(prefixB[i],end=" ")
            print()
    else:
        if(A[0]!=B[-1]):
            ans=len(prefix)+len(prefixB)-1
            print(ans,end=" ")
            for i in range(len(prefix)):
                print(prefix[i],end=" ")
            for i in range(len(prefixB)-2,-1,-1):
                print(prefixB[i],end=" ")
            print()
        else:
            ans=len(prefix)+len(prefixB)
            print(ans,end=" ")
            for i in range(len(prefix)):
                print(prefix[i],end=" ")
            for i in range(len(prefixB)-1,-1,-1):
                print(prefixB[i],end=" ")
            print()