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:
import sys
import math 
from decimal import *
def I(): return int(sys.stdin.readline())
def IL(): return list(map(int,sys.stdin.readline().strip().split()))
def ILS(): return list(map(str,sys.stdin.readline().strip().split()))

def solve():
	tt = I()
	for ii in range(tt):
		n = I()
		a = input()
		b = input()
		# print(a,b,"--")
		a = [int(x) for x in a]
		b = [int(x) for x in b]
		a0=a[0];
		flip=0;
		l=1;r=n-1;
		i = n-1;j = n-1
		ans = []
		while i>=0 :
			flag=0;
			aj=a[j];
			if flip == 1 : 
				if(aj==0) : aj=1;
				else: aj=0;
			
			if(j==0) : aj=a0;
			
			if(aj!=b[i]) : 
				flag=1;
				flip=( not flip);
				if(aj!=a0):
					ans.append(1);
				
				ans.append(i+1);
				if(aj==0) : a0=1;
				else : a0=0;
		 
			if(j==l and  j==r) : j=0;
			elif(j==r):
				r-=1;
				if(flag == 1): j=l;
				else : j=r;
			
			else :
				l+=1;
				if(flag == 1):j=r;
				else: j=l;
			i-=1
			
			
		
		print(len(ans),end=" ")
		
		for xx in  ans:
			print(xx,end=" ")
		print()
		

solve()