#include <cstdio>

#define MAX 150

char sp[MAX][MAX];
//u r d l
const int dx[] = {0,1,0,-1};
const int dy[] = {-1,0,1,0};

bool range(int x,int y,int n){
  if(0 > x || x >= n || 0 > y || y >= n){
    return false;
  }
  return true;
}

void make(int n){
  int x = 0 , y = n-1 ,dir = 0;

  for(int i = 0 ; i < n ; i++){
    for(int j = 0 ; j < n ; j++){
      sp[i][j] = '.';
    }
  }

  int end = 0;
  sp[y][x] = '#';
  while(true){
    int nx = x + dx[dir];
    int ny = y + dy[dir];  

    bool flg = true;
    if(!range(nx,ny,n)){
      flg = false;
    }

    for(int i = 0 ; i < 4 ; i++){
      int nnx = nx + dx[i];
      int nny = ny + dy[i];

      if((x == nnx && y == nny) 
	 || !range(nnx,nny,n)){
	continue;
      }

      if(sp[nny][nnx] == '#'){
	flg = false;
	break;
      }
    }

    if(!flg){
      dir++;
      dir = dir % 4;
      end++;
      if(end == 4) return;
    }else{
      end = 0;
    }

    if(flg){
      sp[ny][nx] = '#';
      x = nx;
      y = ny;
    }
  }
}

int main(){
  int d,n;
  bool ok = false;

  scanf("%d" ,&d);
  while(d--){
    scanf("%d" ,&n);

    if(ok){
      puts("");
    }
    ok = true;

    make(n);
    for(int i = 0 ; i < n ; i++){
      for(int j = 0 ; j < n ; j++){
	if(sp[i][j] == '.'){
	  printf(" ");
	}else{
	  printf("%c" ,sp[i][j]);
	}
      }
      puts("");
    }   
  }

  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 <cstdio>

#define MAX 150

char sp[MAX][MAX];
//u r d l
const int dx[] = {0,1,0,-1};
const int dy[] = {-1,0,1,0};

bool range(int x,int y,int n){
  if(0 > x || x >= n || 0 > y || y >= n){
    return false;
  }
  return true;
}

void make(int n){
  int x = 0 , y = n-1 ,dir = 0;

  for(int i = 0 ; i < n ; i++){
    for(int j = 0 ; j < n ; j++){
      sp[i][j] = '.';
    }
  }

  int end = 0;
  sp[y][x] = '#';
  while(true){
    int nx = x + dx[dir];
    int ny = y + dy[dir];  

    bool flg = true;
    if(!range(nx,ny,n)){
      flg = false;
    }

    for(int i = 0 ; i < 4 ; i++){
      int nnx = nx + dx[i];
      int nny = ny + dy[i];

      if((x == nnx && y == nny) 
	 || !range(nnx,nny,n)){
	continue;
      }

      if(sp[nny][nnx] == '#'){
	flg = false;
	break;
      }
    }

    if(!flg){
      dir++;
      dir = dir % 4;
      end++;
      if(end == 4) return;
    }else{
      end = 0;
    }

    if(flg){
      sp[ny][nx] = '#';
      x = nx;
      y = ny;
    }
  }
}

int main(){
  int d,n;
  bool ok = false;

  scanf("%d" ,&d);
  while(d--){
    scanf("%d" ,&n);

    if(ok){
      puts("");
    }
    ok = true;

    make(n);
    for(int i = 0 ; i < n ; i++){
      for(int j = 0 ; j < n ; j++){
	if(sp[i][j] == '.'){
	  printf(" ");
	}else{
	  printf("%c" ,sp[i][j]);
	}
      }
      puts("");
    }   
  }

  return 0;
}
```