#include <bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define repb(i,a,b) for(int i=a;i>=b;i--)
#define all(a) a.begin(),a.end()
#define o(a) cout<<a<<endl
#define int long long
#define fi first
#define se second
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;

int h, w;

bool can(int y, int x){
	return (0 <= y && y < h && 0 <= x && x < w);
}

int dy[4] = {-1, 0, 1, 0};
int dx[4] = {0, 1, 0, -1};

signed main() {
	int t, n;
	cin >> t;
	rep(k, 0, t){
		if(k) cout << endl;
		int a[110][110]={};
		cin >> n;
		h = w = n;
		vi d;
		int tmp = n;
		d.push_back(tmp);
		while(1){
			if(tmp <= 2){
				d.push_back(tmp);
				break;
			}
			rep(i,0,2) d.push_back(tmp);
			tmp -= 2;
		}
		int y = n - 1, x = 0;
		a[y][x] = 1;
		rep(i, 0, d.size()){
			int dir = i % 4;
			rep(j, 0, d[i] - 1){
				y += dy[dir];
				x += dx[dir];
				a[y][x] = 1;
			}
		}
		rep(i, 0, n){
			rep(j, 0, n){
				cout << (a[i][j] ? "#":" ") ;
			}
			cout << endl;
		}
	}
}
### Prompt

Your task is to create 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 <bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define repb(i,a,b) for(int i=a;i>=b;i--)
#define all(a) a.begin(),a.end()
#define o(a) cout<<a<<endl
#define int long long
#define fi first
#define se second
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;

int h, w;

bool can(int y, int x){
	return (0 <= y && y < h && 0 <= x && x < w);
}

int dy[4] = {-1, 0, 1, 0};
int dx[4] = {0, 1, 0, -1};

signed main() {
	int t, n;
	cin >> t;
	rep(k, 0, t){
		if(k) cout << endl;
		int a[110][110]={};
		cin >> n;
		h = w = n;
		vi d;
		int tmp = n;
		d.push_back(tmp);
		while(1){
			if(tmp <= 2){
				d.push_back(tmp);
				break;
			}
			rep(i,0,2) d.push_back(tmp);
			tmp -= 2;
		}
		int y = n - 1, x = 0;
		a[y][x] = 1;
		rep(i, 0, d.size()){
			int dir = i % 4;
			rep(j, 0, d[i] - 1){
				y += dy[dir];
				x += dx[dir];
				a[y][x] = 1;
			}
		}
		rep(i, 0, n){
			rep(j, 0, n){
				cout << (a[i][j] ? "#":" ") ;
			}
			cout << endl;
		}
	}
}
```