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 collections import deque
from math import *
import heapq
def func(s1,s2):
    return
def negate(s1,ind):
    for i in range(ind+1):
        if s1[i]=='0':
            s1[i]='1'
        else:
            s1[i]='0'
    return s1[:ind+1]
 
t=int(input())
for i in range(t):
    n=int(input())
    S1=input()
    S2=input()
    s1=list(S1)
    s2=list(S2)
    # print(s1)
    # s1[2]="3"
    # print(s1)
 
    if s1==s2:
        print(0)
        continue
    end1=len(s1)-1
    end2=len(s1)-1
    
    ANSWER=[]
 
    while end1>=0:
        if s1[end1]==s2[end2]:
            end1-=1
            end2-=1
    
        else:
            if s1[0]!=s2[end2]:
                s1[:end1+1]=negate(s1,end1)
                s1[:end1+1]=s1[:end1+1][-1::-1]
                ANSWER.append(end1+1)
                
            else:
                ANSWER.append(1)
                s1[0]=negate(s1,0)
 
                s1[:end1+1]=negate(s1,end1)
                s1[:end1+1]=s1[:end1+1][-1::-1]
                ANSWER.append(end1+1)
                
                
            end1-=1
            end2-=1
        # print(s1,ANSWER,end1)
    print(len(ANSWER),*ANSWER)