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:
######################################################################
# Write your code here
import sys
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x10000000, resource.RLIM_INFINITY])
#sys.setrecursionlimit(0x100000)
# Write your code here
# For getting input from input.txt file 
# sys.stdin = open('input.txt', 'r')  
# Printing the Output to output.txt file 
# sys.stdout = open('output.txt', 'w')
RI =   lambda : [int(x) for x in sys.stdin.readline().strip().split()]
rw =   lambda : sys.stdin.readline().strip().split()
ri =   lambda : int(sys.stdin.readline().strip())
rs =   lambda : sys.stdin.readline().strip()
from collections import defaultdict as df
#import heapq 
#heapq.heapify(li) heappush(li,4) heappop(li)
#import random
#random.shuffle(list)
#infinite = float('inf')
#######################################################################
# from queue import PriorityQueue as pq
from math import *

t = ri()
for _ in range(t):
    n = ri()
    l = list(rs())
    l = [int(i) for i in l]
    m = list(rs())
    m = [int(i) for i in m]

    # print(l)
    # print(m)
    direction = 1
    ans = []
    left = 0
    right = n-1
    bit = 0#invert
    i = n-1
    while(i>=0 and right>=left):
        if(direction==1):
            
            # l[right] = l[right]^bit

            if(l[right]^bit==m[i]):
                i-=1
                right-=1
                continue
            if(l[left]!=l[right]):
                ans.append(1)
                l[left]^=1

                ans.append(right-left+1)
                bit^=1#inverted

                direction^=1#left se
                # l = list(reversed(l[:i+1]))#+l[i+1:]
                # print(l)
                left+=1
                i-=1
            else:
                ans.append(right-left+1)
                bit^=1#inverted

                direction^=1#left se
                # l = list(reversed(l[:i+1]))#+l[i+1:]
                # print(l)
                left+=1
                i-=1
        else:
            # l[left] = l[left]^bit

            if(l[left]^bit==m[i]):
                i-=1
                left+=1
                continue
            if(l[left]!=l[right]):
                ans.append(1)
                l[right]^=1

                ans.append(right-left+1)
                bit^=1#inverted

                direction^=1#left se
                # l = list(reversed(l[:i+1]))#+l[i+1:]
                # print(l)
                right-=1
                i-=1
            else:
                ans.append(right-left+1)
                bit^=1#inverted

                direction^=1#left se
                # l = list(reversed(l[:i+1]))#+l[i+1:]
                # print(l)
                right-=1
                i-=1
    print(len(ans),*ans)