A1. Prefix Flip (Easy Version)

A1. Prefix Flip (Easy Version)

A1. Prefix Flip (Easy Version)

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

This is the easy version of the problem. The difference between the versions is the constraint on nn and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings aa and bb of length nn (a binary string is a string consisting of symbols 00 and 11). In an operation, you select a prefix of aa, and simultaneously invert the bits in the prefix (00 changes to 11 and 11 changes to 00) and reverse the order of the bits in the prefix.

For example, if a=001011a=001011 and you select the prefix of length 33, it becomes 011011011011. Then if you select the entire string, it becomes 001001001001.

Your task is to transform the string aa into bb in at most 3n3n operations. It can be proved that it is always possible.

Input

The first line contains a single integer tt (1≤t≤10001≤t≤1000)  — the number of test cases. Next 3t3t lines contain descriptions of test cases.

The first line of each test case contains a single integer nn (1≤n≤10001≤n≤1000)  — the length of the binary strings.

The next two lines contain two binary strings aa and bb of length nn.

It is guaranteed that the sum of nn across all test cases does not exceed 10001000.

Output

For each test case, output an integer kk (0≤k≤3n0≤k≤3n), followed by kk integers p1,…,pkp1,…,pk (1≤pi≤n1≤pi≤n). Here kk is the number of operations you use and pipi is the length of the prefix you flip in the ii-th operation.

Example

input

Copy

5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1

output

Copy

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→1001→11→00→10.

In the second test case, we have 01011→00101→11101→01000→10100→00100→1110001011→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 22, which will leave aa unchanged.

=========================================================================

类似问题做过多次，往往就是针对一位置不停改变，使得其余位置不受影响，突破点在于给定的次数，3*n，也就暗示着一个点进行三次变换可以达到只修改着一个点，其余点不会发生改变

XXXXXXXX1把最后一个1变成0而其余不受影响可以按照一下三步

翻转

1XXXXXXXX

翻转1

0XXXXXXXX

打回去

XXXXXXXX1

正好3步

# include<iostream>
# include<string.h>
# include<vector>

using namespace std;

string s,t;
int n;
vector<int>v;
int main ()
{

      cin>>n;
      while(n--)
      {
          int len;
          cin>>len;
          cin>>s>>t;

          for(int i=len-1;i>=0;i--)
          {
              if(s[i]!=t[i])
              {
                  v.push_back(i+1);
              }
          }

          cout<<v.size()*3<<endl;

          for(auto it:v)
          {
              cout<<it<<" "<<1<<" "<<it<<" ";
          }
          cout<<endl;
          v.clear();
      }



    return  0;

}