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.
z=input
mod = 10**9 + 7
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
def lcd(xnum1,xnum2):
    return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
        p=ceil(x**.5)+1
        for i in range(2,p):
            if x%i==0:
                return 0
        return 1

################################################################################

"""

n=int(z())

for _ in range(int(z())):

x=int(z())

l=list(map(int,z().split()))

n=int(z())

l=sorted(list(map(int,z().split())))[::-1]

a,b=map(int,z().split())

l=set(map(int,z().split()))

led=(6,2,5,5,4,5,6,3,7,6)

vowel={'a':0,'e':0,'i':0,'o':0,'u':0}

color4=["G", "GB", "YGB", "YGBI", "OYGBI" ,"OYGBIV",'ROYGBIV' ]

"""

###########################---START-CODING---###############################################
num=int(z())

for _ in range(num):
    n=int(z())
    l1=list(z())+['0']
    l2=list(z())+['0']
    lp,lx=[],[]
    for i in range(1,n+1):
        if l1[i]!=l1[i-1]:
            lp.append(i)
        if l2[i]!=l2[i-1]:
            lx.append(i)
    print(len(lp)+len(lx),*lp,*lx[::-1])
        
            
    
        
        
            
            
            
        
            

    
        
            
            
        
    
        
        
