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, math
input = sys.stdin.readline

def getInts():
    return [int(s) for s in input().split()]

def getInt():
    return int(input())

def getStrs():
    return [s for s in input().split()]

def getStr():
    return input().strip()

def listStr():
    return list(input().strip())

import collections as col
import math

def rev(arr):
    return arr[::-1]

def comp(arr):
    return [arr[j]^1 for j in range(len(arr))]
    
def RC(arr):
    return rev(comp(arr))
    
"""
01
01

"""

def solve():
    N = getInt()
    S1 = [int(s) for s in input().strip()]
    S2 = [int(s) for s in input().strip()]
    index = N-1
    ans = []
    parity = 1
    curr_max = N-1
    curr_min = 0
    checker = []
    #if parity is even, curr_min is at the front, else curr_max is at the front
    while index > -1:
        targ = S2[index]
        ref = S1[curr_min] if parity > 0 else 1-S1[curr_max]
        #print("CRT",curr_min,ref,targ)
        if index == 0:
            if ref != S2[index]:
                ans.append(1)
                if parity > 0:
                    S1[curr_min] = 1-S1[curr_min]
                else:
                    S1[curr_max] = 1-S1[curr_max]
        elif ref != targ:
            ans.append(index+1)
        else:
            ans.append(1)
            if parity > 0:
                S1[curr_min] = 1-S1[curr_min]
            else:
                S1[curr_max] = 1-S1[curr_max]
            ans.append(index+1)
        parity *= -1
        index -= 1
        if parity < 0:
            curr_min += 1
        else:
            curr_max -= 1
        # if index > 0:
            # print("CM",curr_max,curr_min)
    print(len(ans),*ans)
    return

"""
5
2
01
10
3 1 2 1
5
01011
11100
8 1 5 1 4 1 3 2 1
2
01
01
1 2
10
0110011011
1000110100
19 1 10 1 9 1 8 1 7 1 6 1 5 1 4 1 3 1 2 1
1
0
1
1 1
"""
# arr = [j for j in range(10)]
# print(arr)
# for j in range(10):
    # arr = rev(arr[:10-j])+arr[10-j:]
    # print(arr)
        
for _ in range(getInt()):
    solve()