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.
'''from heapq import * 
t = int(input())
for _ in range(t):
    n,w = map(int,input().split())
    a = list(map(int,input().split()))
    a.sort(reverse=True)
    h = [0]
    for i in range(n):
        t =heappop(h)
        if t+a[i]<=w:
            heappush(h,t+a[i])
        else:
            heappush(h,a[i])
            heappush(h,t)
    print(len(h))    '''    
t = int(input())
for _ in range(t):
    n = int(input())
    a = input()
    b = input()
    a = list(a)
    b = list(b)
    k = 0
    i = 0
    l = []
    j = 0
    while i<n and j<n:
        if a[i]==b[j]:
            i+=1
            j+=1
        else:
            k+=3
            l.append(i+1)
            for x in range(i+1):
                if a[x]=='1':
                    a[x]='0'
                else:
                    a[x]='1'
            c = a[0:i+1]
            c.reverse()
            a[0:i+1]=c
            if a[0]=='1':
                a[0]=='0'
            else:
                a[0]='1'
            l.append(1)
            for x in range(i+1):
                if a[x]=='1':
                    a[x]='0'
                else:
                    a[x]='1'
            c = a[0:i+1]
            c.reverse()
            a[0:i+1]=c    
            l.append(i+1)
            i+=1
            j+=1
            
    if k==0:
        print(0)
    else:
        print(k,end=' ')
        for x in range(len(l)):
            print(l[x],end= ' ')
            
    print(' ')    
                  
            
            
            
        
        
  
    
    
        