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 bisect

import math

def st():
    return list(stdin.readline().strip())

def inp():
    return int(stdin.readline())

def li():
    return list(map(int,stdin.readline().split()))

def mp():
    return map(int,stdin.readline().split())

def pr(n):
    stdout.write(str(n)+"\n")

def soe(limit):
    l=[1]*(limit+1)
    prime=[]
    for i in range(2,limit+1):
        if l[i]:
            for j in range(i*i,limit+1,i):
                l[j]=0

    for i in range(2,limit+1):
        if l[i]:
            prime.append(i)
    return prime

def segsoe(low,high):
    limit=int(high**0.5)+1
    prime=soe(limit)
    n=high-low+1
    l=[0]*(n+1)
    for i in range(len(prime)):
        lowlimit=(low//prime[i])*prime[i]
        if lowlimit<low:
            lowlimit+=prime[i]
        if lowlimit==prime[i]:
            lowlimit+=prime[i]
        for j in range(lowlimit,high+1,prime[i]):
            l[j-low]=1
    for i in range(low,high+1):
        if not l[i-low]:
            if i!=1:
                print(i)

def power(a,n):
    r=1
    while n:
        if n&1:
            r=(r*a)
        a*=a
        n=n>>1
    return r


def solve():
    n=inp()
    a=input()
    b=input()
    j=n-1
    x=[]
    if a==b:
        print(0)
        return
    while j>=0:
        if a[j]==b[j]:
            j-=1
            
        elif a[0]!=b[j]:
            x.append(j+1)
            p=''
            for i  in range(j+1):
                if a[i]=='1':
                    p+='0'
                else:
                    p+='1'
            a=p[::-1]+a[j:]
            j-=1
        else:
            x.append(1)
            if a[0]=='1':
                a='0'+a[1:]
            else:
                a='1'+a[1:]
 
    print(len(x),*x)
 
for _ in range(inp()):
    solve()