#include <iostream>
#include <vector>

using namespace std;

int main(void){
  int d;cin>>d;
  const int POS[][2]={{0,-1},{1,0},{0,1},{-1,0}};
  while(d--){
    int n;cin>>n;
    vector<vector<char> > field(n+2);
    for(int i=0;i<=n+1;++i){
      field[i].reserve(n+2);
      for(int j=0;j<=n+1;++j)field[i][j]=(i==n+1||j==n+1)?1:0;
    }
    int x=0,y=n-1,p=0;
    for(;;){
    	if(x<0||y<0)break;
      int dx=POS[p][0],dy=POS[p][1];
      int px=POS[(p+4+1)%4][0],py=POS[(p+4+1)%4][1],qx=POS[(p+4-1)%4][0],qy=POS[(p+4-1)%4][1];
      field[y][x]=1;
      x+=dx,y+=dy;
      if(y==0&&x==0||(y+2*dy>=0&&x+2*dx>=0&&field[y+2*dy][x+2*dx]==1))p=(p+1)%4;
      if((y+dy+py>=0&&x+dx+px>=0&&field[y+dy+py][x+dx+px]==1)||(y+dy+qy>=0&&x+dx+qx>=0&&field[y+dy+qy][x+dx+qx]==1))
      	break;
    }
    if(n==2)field[1][1]=0;
    for(int i=0;i<n;++i){
      for(int j=0;j<n;++j)cout<<((field[i][j])?'#':' ');
      cout<<endl;
    }
    if(d)cout<<endl;
  }
  return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

I decided to create a program that displays a "round and round pattern". The "round and round pattern" is as follows.

* If the length of one side is n, it is displayed as a character string with n rows and n columns.
* A spiral pattern that rotates clockwise with the lower left corner as the base point.
* The part with a line is represented by # (half-width sharp), and the blank part is represented by "" (half-width blank).
* Leave a space between the lines.



Create a program that takes an integer n as an input and outputs a "round and round pattern" with a side length of n.



Input

The input is given in the following format:


d
n1
n2
::
nd


The number of datasets d (d ≤ 20) is given to the first line, and the side length ni (1 ≤ ni ≤ 100) of the i-th round pattern is given to each of the following d lines.

Output

Please output a round and round pattern for each data set. Insert a blank line between the datasets.

Example

Input

2
5
6


Output

#####
#   #
# # #
# # #
# ###

######
#    #
# ## #
# #  #
# #  #
# ####

### Response

```cpp
#include <iostream>
#include <vector>

using namespace std;

int main(void){
  int d;cin>>d;
  const int POS[][2]={{0,-1},{1,0},{0,1},{-1,0}};
  while(d--){
    int n;cin>>n;
    vector<vector<char> > field(n+2);
    for(int i=0;i<=n+1;++i){
      field[i].reserve(n+2);
      for(int j=0;j<=n+1;++j)field[i][j]=(i==n+1||j==n+1)?1:0;
    }
    int x=0,y=n-1,p=0;
    for(;;){
    	if(x<0||y<0)break;
      int dx=POS[p][0],dy=POS[p][1];
      int px=POS[(p+4+1)%4][0],py=POS[(p+4+1)%4][1],qx=POS[(p+4-1)%4][0],qy=POS[(p+4-1)%4][1];
      field[y][x]=1;
      x+=dx,y+=dy;
      if(y==0&&x==0||(y+2*dy>=0&&x+2*dx>=0&&field[y+2*dy][x+2*dx]==1))p=(p+1)%4;
      if((y+dy+py>=0&&x+dx+px>=0&&field[y+dy+py][x+dx+px]==1)||(y+dy+qy>=0&&x+dx+qx>=0&&field[y+dy+qy][x+dx+qx]==1))
      	break;
    }
    if(n==2)field[1][1]=0;
    for(int i=0;i<n;++i){
      for(int j=0;j<n;++j)cout<<((field[i][j])?'#':' ');
      cout<<endl;
    }
    if(d)cout<<endl;
  }
  return 0;
}
```