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 sys import stdin
input = stdin.readline

for _ in range(int(input())):
    n = int(input())
    a = input().strip()
    b = input().strip()
    bi = n-1
    f = 0
    l, r = 0, n-1
    ans = []
    #temp = list(a)
    while l!=r:
        #print(l,r)
        if (f==0 and a[l]==b[bi]) or (f!=0 and a[l]!=b[bi]):
            #temp[0] = '0' if temp[0]=='1' else '1'
            ans.append(1)
        #for i in range(0, abs(l-r)+1):
        #    temp[i] = '0' if temp[i]=='1' else '1'
        #print(temp,'mid')
        #temp[0: abs(l-r)+1] = temp[0: abs(l-r)+1][::-1]
        ans.append(abs(l-r)+1)
        if l<r:
            l += 1
        else:
            l -= 1
        l, r = r, l
        f ^= 1
        bi-=1
        #print(temp)
    if (f==0 and a[l]!=b[0]) or (f and a[l]==b[0]):
        #temp[0] = '0' if temp[0]=='1' else '1'
        ans.append(1)
    
    print(len(ans),end=' ')
    print(*ans)
    #print(temp)