#include <stdio.h>
#pragma warning(disable : 4996)
int n, d, x, y, e, f; char c[100][101];
const int dx[4] = { 0, 1, 0, -1 };
const int dy[4] = { -1, 0, 1, 0 };
bool ok() {
	for (int i = 0; i < 4; i++) {
		if (i == (e + 2) % 4) continue;
		int ex = x + dx[i];
		int ey = y + dy[i];
		if (0 <= ex && ex < d && 0 <= ey && ey < d) {
			if (c[ey][ex] == '#') {
				return false;
			}
		}
	}
	return true;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &d);
		x = 0, y = d - 1, e = 0, f = 0;
		for (int j = 0; j < d; j++) {
			for (int k = 0; k < d; k++) {
				c[j][k] = ' ';
			}
			c[j][d] = '\0';
		}
		while (f++ < d * d) {
			c[y][x] = '#';
			x += dx[e % 4];
			y += dy[e % 4];
			if (0 <= x && x < d && 0 <= y && y < d) {
				if (!ok()) {
					x -= dx[e % 4];
					y -= dy[e % 4];
					e++;
				}
			}
			else {
				x -= dx[e % 4];
				y -= dy[e % 4];
				e++;
			}
		}
		if (i) printf("\n");
		for (int j = 0; j < d; j++) printf("%s\n", c[j]);
	}
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <stdio.h>
#pragma warning(disable : 4996)
int n, d, x, y, e, f; char c[100][101];
const int dx[4] = { 0, 1, 0, -1 };
const int dy[4] = { -1, 0, 1, 0 };
bool ok() {
	for (int i = 0; i < 4; i++) {
		if (i == (e + 2) % 4) continue;
		int ex = x + dx[i];
		int ey = y + dy[i];
		if (0 <= ex && ex < d && 0 <= ey && ey < d) {
			if (c[ey][ex] == '#') {
				return false;
			}
		}
	}
	return true;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &d);
		x = 0, y = d - 1, e = 0, f = 0;
		for (int j = 0; j < d; j++) {
			for (int k = 0; k < d; k++) {
				c[j][k] = ' ';
			}
			c[j][d] = '\0';
		}
		while (f++ < d * d) {
			c[y][x] = '#';
			x += dx[e % 4];
			y += dy[e % 4];
			if (0 <= x && x < d && 0 <= y && y < d) {
				if (!ok()) {
					x -= dx[e % 4];
					y -= dy[e % 4];
					e++;
				}
			}
			else {
				x -= dx[e % 4];
				y -= dy[e % 4];
				e++;
			}
		}
		if (i) printf("\n");
		for (int j = 0; j < d; j++) printf("%s\n", c[j]);
	}
	return 0;
}
```