#include<iostream>
#include<cassert>
#include<ctime>
#include<climits>
#include<queue>
#include<fstream>
#include<deque>
#include<iomanip>
#include<list>
#include<stack>
#include<sstream>
#include<string>
#include<cctype>
#include<cstdio>
#include<vector>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<map>
#include<set>
#define REP(i,s,n) for(int i=s;i<n;i++)
#define rep(i,n) REP(i,0,n)
#define EPS 1e-10
#define F first
#define S second
#define all(n) n.begin(),n.end()
#define pb push_back
#define insert(a,b,c,d) PP(P(a,b),P(c,d))
#define INF (1<<27)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<P,P> PP;
const int dx[4] = {-1,0,+1,0};
const int dy[4] = {0,-1,0,+1};

class Solve{
private:
  int d;
  int imap[110][110];
public:

  Solve(int n):d(n){}
  ~Solve(){}

  int Connect(P p){
    int cnt = 0;
    rep(i,4){
      int nx = p.F+dx[i],ny = p.S+dy[i];
      if(!(0<=nx && nx <d && 0<= ny && ny < d))continue;
      if(imap[ny][nx] == 1)cnt++;
    }
    return cnt;
  }


  void display(){
      cout << "--- in display ---" << endl;
      rep(i,d){
	rep(j,d){
	  if(imap[i][j] == 1)cout << '#';
	  else cout << ' ';
	}
	cout << endl;
      }
      cout << endl;
    }


  void run(){
    rep(i,d+1)rep(j,d+1)imap[i][j] = 0;

    rep(i,d){
      imap[0][i] = 1;
      imap[i][0] = 1;
      imap[i][d-1] = 1;
    }

  


    P pos = P(d-1,d-1);
    int dir = 0;
    int cnt = 0;
    bool fin = false;
    while(true){

      int nx,ny;
      nx  = pos.F,ny = pos.S;
      while(true){
	//cout << "nx = " << nx << ",ny = "<< ny << ",dir = " << dir << endl;
	//display();


	if(!(Connect(P(nx,ny)) == 2 || Connect(P(nx,ny)) == 1)){
	  imap[ny-dy[dir]][nx-dx[dir]] = 0;
	  //cout << "Lase nx = " << nx << ",ny = "<< ny << endl;
	  //cout << "FIN! connnect = " << Connect(P(nx,ny)) << endl;
	  fin = true;
	  break;
	}

	if(imap[ny+dy[dir]][nx+dx[dir]] == 1){
	  //cout << "IN nx = " << nx+dx[dir] << ",ny = " << ny+dy[dir] << endl;
	  imap[ny][nx] = 0;
	  cnt--;
	  break;
	}
	imap[ny][nx] = 1;
	cnt++;	


	nx += dx[dir],ny += dy[dir];
      }
      //cout << "------ cnt = " << cnt << endl;
      if(fin){
	//cout << "dir = " << dir << endl;
	break;
      }
      if(cnt == 0)break;
      cnt =0;

      pos.F = nx-dx[dir],pos.S = ny-dy[dir];
      dir++;
      dir = dir%4;
    }

    rep(i,d){
      rep(j,d){
	if(imap[i][j] == 1)cout << '#';
	else cout << ' ';
      }
      cout << endl;
    }
    

  }


};


int main(){
  int T;
  cin >> T;
  while(T-- > 0){
    int _n;
    cin >> _n;
    Solve ans = Solve(_n);
    ans.run();
    if(T != 0)cout << endl;
  }
return 0;
}
### Prompt

Your challenge is to write a cpp solution to 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<cassert>
#include<ctime>
#include<climits>
#include<queue>
#include<fstream>
#include<deque>
#include<iomanip>
#include<list>
#include<stack>
#include<sstream>
#include<string>
#include<cctype>
#include<cstdio>
#include<vector>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<map>
#include<set>
#define REP(i,s,n) for(int i=s;i<n;i++)
#define rep(i,n) REP(i,0,n)
#define EPS 1e-10
#define F first
#define S second
#define all(n) n.begin(),n.end()
#define pb push_back
#define insert(a,b,c,d) PP(P(a,b),P(c,d))
#define INF (1<<27)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<P,P> PP;
const int dx[4] = {-1,0,+1,0};
const int dy[4] = {0,-1,0,+1};

class Solve{
private:
  int d;
  int imap[110][110];
public:

  Solve(int n):d(n){}
  ~Solve(){}

  int Connect(P p){
    int cnt = 0;
    rep(i,4){
      int nx = p.F+dx[i],ny = p.S+dy[i];
      if(!(0<=nx && nx <d && 0<= ny && ny < d))continue;
      if(imap[ny][nx] == 1)cnt++;
    }
    return cnt;
  }


  void display(){
      cout << "--- in display ---" << endl;
      rep(i,d){
	rep(j,d){
	  if(imap[i][j] == 1)cout << '#';
	  else cout << ' ';
	}
	cout << endl;
      }
      cout << endl;
    }


  void run(){
    rep(i,d+1)rep(j,d+1)imap[i][j] = 0;

    rep(i,d){
      imap[0][i] = 1;
      imap[i][0] = 1;
      imap[i][d-1] = 1;
    }

  


    P pos = P(d-1,d-1);
    int dir = 0;
    int cnt = 0;
    bool fin = false;
    while(true){

      int nx,ny;
      nx  = pos.F,ny = pos.S;
      while(true){
	//cout << "nx = " << nx << ",ny = "<< ny << ",dir = " << dir << endl;
	//display();


	if(!(Connect(P(nx,ny)) == 2 || Connect(P(nx,ny)) == 1)){
	  imap[ny-dy[dir]][nx-dx[dir]] = 0;
	  //cout << "Lase nx = " << nx << ",ny = "<< ny << endl;
	  //cout << "FIN! connnect = " << Connect(P(nx,ny)) << endl;
	  fin = true;
	  break;
	}

	if(imap[ny+dy[dir]][nx+dx[dir]] == 1){
	  //cout << "IN nx = " << nx+dx[dir] << ",ny = " << ny+dy[dir] << endl;
	  imap[ny][nx] = 0;
	  cnt--;
	  break;
	}
	imap[ny][nx] = 1;
	cnt++;	


	nx += dx[dir],ny += dy[dir];
      }
      //cout << "------ cnt = " << cnt << endl;
      if(fin){
	//cout << "dir = " << dir << endl;
	break;
      }
      if(cnt == 0)break;
      cnt =0;

      pos.F = nx-dx[dir],pos.S = ny-dy[dir];
      dir++;
      dir = dir%4;
    }

    rep(i,d){
      rep(j,d){
	if(imap[i][j] == 1)cout << '#';
	else cout << ' ';
      }
      cout << endl;
    }
    

  }


};


int main(){
  int T;
  cin >> T;
  while(T-- > 0){
    int _n;
    cin >> _n;
    Solve ans = Solve(_n);
    ans.run();
    if(T != 0)cout << endl;
  }
return 0;
}
```