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:
t=int(input())
for i in range(t):
    n=int(input())
    x=input()
    y=input()
    a=[]
    b=[]
    for j in x:
        a.append(j)
    for j in y:
        b.append(j)
    p=[]
    j=0
    while(j<n):
        if a[j]!=b[j]:
            if j==(n-1):
                if j!=0:
                    p.append(j + 1)
                    p.append(1)
                    p.append(j + 1)
                else:
                    p.append(1)
                j+=1

            else:
                if a[j + 1] == b[j + 1]:
                    if j!=0:
                        p.append(j + 1)
                        p.append(1)
                        p.append(j + 1)
                    else:
                        p.append(1)

                    j += 2
                elif a[j+1]!=b[j+1]:
                    if a[j]==a[j+1]:
                        p.append(j+2)
                        p.append(2)
                        p.append(j+2)
                        j+=2
                    else:
                        if j==(n-2):
                            if j!=0:
                                p.append(j + 2)
                                p.append(1)
                                p.append(2)
                                p.append(1)
                                p.append(j + 2)
                            else:
                                p.append(1)
                                p.append(2)
                                p.append(1)

                            j+=2
                        else:
                            if a[j+2]==b[j+2]:
                                p.append(j + 2)
                                p.append(1)
                                p.append(2)
                                p.append(1)
                                p.append(j + 2)
                                j+=3
                            else:
                                if a[j]==a[j+2]:
                                    p.append(j+3)
                                    p.append(3)
                                    p.append(j+3)
                                    j+=3
                                elif a[j+1]==a[j+2]:
                                    if j!=0:
                                        p.append(j + 1)
                                        p.append(1)
                                        p.append(j + 1)
                                    else:
                                        p.append(1)

                                    p.append(j + 3)
                                    p.append(2)
                                    p.append(j + 3)
                                    j+=3
        else:
            j+=1




    print(len(p),end=" ")
    print(*p)