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.
import sys
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
from copy import *
from bisect import *	
from heapq import *
from math import *
from itertools import permutations 
 
def eprint(*args):
    print(*args, file=sys.stderr)
zz=1
 
#sys.setrecursionlimit(10**6)
if zz:
	input=sys.stdin.readline
else:	
	sys.stdin=open('input.txt', 'r')
	sys.stdout=open('all.txt','w')
def li():
	return [int(xx) for xx in input().split()]
def fli():
	return [float(x) for x in input().split()]	
def comp(a,b):
	if(a>b):
		return 2
	return 2 if a==b else 0		
def gi():	
	return [xx for x in input().split()]
def fi():
	return int(input())
def swap(a,i,j):
	a[i],a[j]=a[j],a[i]	
def si():
	return list(input().rstrip())	
def mi():
	return 	map(int,input().split())	
def gh():
	sys.stdout.flush()
def graph(n,m):
	for i in range(m):
		x,y=mi()
		a[x].append(y)
		a[y].append(x)
def bo(i):
	return ord(i)-ord('a')
		


t=fi()

while t>0:
    t-=1
    n=fi()
    s=si()
    l=si()
    for i in range(n):
    	s[i]=int(s[i])
    	l[i]=int(l[i])
    d=[]
    #0->5->1->4->2->3
    for i in range(n):
    	if i%2==0:
    		if l[n-i-1]==s[i//2]:
    			d.append(1)
    			s[i//2]^=1
    		d.append(n-i)
    	else:
    		s[n-(i+1)//2]^=1
    		if l[n-i-1]==s[n-(i+1)//2]:
    			d.append(1)
    			s[n-(i+1)//2]^=1
    		d.append(n-i)

    print(len(d),*d)	
						