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=input('')
    a=list(map(int,a))
    b=input('')
    b=list(map(int,b))
    if a==b:
        print(0)
        continue
    def reverse(n):
        for i in range(n//2):
            a[i]^=1
            a[n-i-1]^=1
            a[i],a[n-i-1]=a[n-i-1],a[i]
        if n&1:
            a[n//2]^=1

    ans=[]
    count=0
    flip=0
    last=n-1
    start=0
    index=0
    j=n-1
    while(start<=last and j>=0):
        if index%2==0:
            if flip&1:
                a[start]^=1
            if a[start]==b[j]:
                count+=1
                a[start]^=1
                ans.append(1)
            count+=1
            flip+=1
            ans.append(j+1)
            start+=1
        else:
            if flip&1:
                a[last]^=1
            if a[last]==b[j]:
                count+=1
                a[last]^=1
                ans.append(1)
            count+=1
            flip+=1
            ans.append(j+1)
            last-=1
        j-=1
        index+=1

    print(count,*ans)