#include<iostream>
#include<string>
using namespace std;
int main(){
long d,n,i,j,k,p,q,r;
string s[100];
cin>>d;
for(k=0;k<d;k++){
cin>>n;
for(i=0;i<n;i++){
s[i].erase();
s[i].resize(n,' ');
}
for(i=n-1;i>=0; i--)s[i][0]='#';
for(j=0;j<n/2;j+=2){
for(i=j;    i<n-j;  i++)s[j]    [i]    ='#';
if(2*j>=n-2)break;
for(i=j;    i<n-j;  i++)s[i]    [n-1-j]='#';
for(i=n-1-j;i>=j+2; i--)s[n-1-j][i]    ='#';
if(2*j>=n-4)break;
for(i=n-1-j;i>=j+2; i--)s[i]    [j+2]  ='#';
}
if(k>0)cout<<endl;
for(i=0;i<n;i++)cout<<s[i]<<endl;
}
return 0;
}
### Prompt

Create a solution in Cpp for the following problem:

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<string>
using namespace std;
int main(){
long d,n,i,j,k,p,q,r;
string s[100];
cin>>d;
for(k=0;k<d;k++){
cin>>n;
for(i=0;i<n;i++){
s[i].erase();
s[i].resize(n,' ');
}
for(i=n-1;i>=0; i--)s[i][0]='#';
for(j=0;j<n/2;j+=2){
for(i=j;    i<n-j;  i++)s[j]    [i]    ='#';
if(2*j>=n-2)break;
for(i=j;    i<n-j;  i++)s[i]    [n-1-j]='#';
for(i=n-1-j;i>=j+2; i--)s[n-1-j][i]    ='#';
if(2*j>=n-4)break;
for(i=n-1-j;i>=j+2; i--)s[i]    [j+2]  ='#';
}
if(k>0)cout<<endl;
for(i=0;i<n;i++)cout<<s[i]<<endl;
}
return 0;
}
```