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,stdout 
import math,queue,heapq
fastinput=stdin.readline
fastout=stdout.write
t=int(fastinput())
def changeflag(flag):
    return 0 if flag==1 else 1
def original(c,flag):
    if flag==1:
        return c 
    else:
        return '0' if c=='1' else '1'
def ismatch(c,v,flag):
    c=original(c,flag) 
    if c==v:
        return True 
    else :
        return False 

while t:
    t-=1 
    n=int(fastinput()) 
    a=input()
    b=input() 
    start=0 
    end=n-1 
    flag=1 
    instruction=[] 
    for i in range(n-1,-1,-1):
        if start==end:
            if ismatch(a[start],b[i],flag):
                continue 
            else:
                instruction.append(1) 
        elif start<end:
            if ismatch(a[end],b[i],flag):
                end-=1 
            else:
                if ismatch(a[start],b[i],flag):
                    instruction.append(1) 
                flag=changeflag(flag) 
                instruction.append(end-start+1)
                temp=start 
                start=end 
                end=temp 
                end+=1 
        else:
            
            if ismatch(a[end],b[i],flag):
                end+=1 
            else:
                if ismatch(a[start],b[i],flag):
                    instruction.append(1) 
                instruction.append(start-end+1)
                flag=changeflag(flag) 
                temp=start 
                start=end 
                end=temp 
                end-=1 
    print(len(instruction),*instruction)