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:
def examA():
    T = I()
    for _ in range(T):
        N, M = LI()
        A = LI()
        B = LI()
        B = set(B)
        cur = -1
        for a in A:
            if a in B:
                cur = a
                break
        if cur==-1:
            print("NO")
            continue
        print("YES")
        print("1",cur)
    return

def examB():
    T = I()
    ans = []
    for _ in range(T):
        N = I()
        A = LI()
        cur = -1
        for i in range(N):
            if A[i]>1:
                if i%2==0:
                    cur = 1
                else:
                    cur = 0
                break
        if cur==-1:
            if N%2==1:
                cur = 1
            else:
                cur = 0

        if cur==0:
            print("Second")
        else:
            print("First")
    return

def examC():
    T = I()
    for _ in range(T):
        N = I()
        a = SI()
        b = SI()
        A = []
        B = []
        for i in a:
            A.append(int(i))
        for i in b:
            B.append(int(i))
        cur = []
        for l in range(N-1,-1,-1):
            if A[l] == B[l]:
                continue
            if A[0] == B[l]:
                A[0] = 1^A[0]
                cur.append(1)
            ne_A = []
            for i in A[:l+1]:
                ne_A.append(1^i)
            A = ne_A[::-1] + A[l+1:]
            cur.append(l + 1)

        print(len(cur), " ".join(map(str,cur)))
    return

def examD():
    ans = 0
    print(ans)
    return

def examE():
    ans = 0
    print(ans)
    return

def examC2():
    T = I()
    for _ in range(T):
        N = I()
        a = SI()
        b = SI()
        A = []
        B = []
        for i in a:
            A.append(int(i))
        for i in b:
            B.append(int(i))
        cur = []
        rev = 0
        L = 0; R = N-1
        for l in range(N-1,-1,-1):
            if rev==0:
                if A[R] == B[l]:
                    R -= 1
                    continue
                if A[L] == B[l]:
                    A[L] = 1 ^ A[L]
                    cur.append(1)
                cur.append(l + 1)
                L += 1
            else:
                if A[L] != B[l]:
                    L += 1
                    continue
                if A[R] != B[l]:
                    A[R] = 1 ^ A[R]
                    cur.append(1)
                cur.append(l + 1)
                R -= 1
            rev ^= 1

        print(len(cur), " ".join(map(str,cur)))
    return

from decimal import getcontext,Decimal as dec
import sys,bisect,heapq
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = dec("0.000000000001")
alphabet = [chr(ord('a') + i) for i in range(26)]
alphabet_convert = {chr(ord('a') + i): i for i in range(26)}

getcontext().prec = 28

sys.setrecursionlimit(2*10**5)

if __name__ == '__main__':
    examC2()

"""
3 1 2 1
3 1 5 2
0 
7 1 10 8 7 1 2 1

1 1
"""