Lusrica designs a mosquito coil in a board with n × n grids. The mosquito coil is a series of consecutive grids, each two neighboring grids of which share a common border. If two grids in the mosquito coil are not consecutive, they do not share any border, but they can share a common endpoint.

The mosquito coil Lusrica designed starts from the upper left corner of the board. It goes right to the last available grid. Alter the direction and go downward to the last available grid, and alter the direction again going left to the last available grid. To carry on after altering the direction and go upward to the last available grid. Then it goes right again and repeats the above turns.

It ends up in a grid such that the above process cannot be continued any more. Your mission now is to print the whole blueprint of Lusrica's mosquito coil.

Input

This problem has several test cases and the first line contains an integer t (1 ≤ t ≤ 36) which is the number of test cases. For each case a line contains an integer n (1 ≤ n ≤ 36) indicating the size of the board.

Output

For each case with input n, output n lines to describe the whole board. Each line contains n characters. If a grid is a part of Lusrica's mosquito coil, the corresponding character is '#', or ' ' (a single blank) if not.

樣例輸入

5
1
2
3
4
5

樣例輸出

#
##
 #
###
  #
###
####
   #
#  #
####
#####
    #
### #
#   #
#####

青島現場賽題

題目說是一盤蚊香 我覺得有點像貪吃蛇的意思 要碰到自己的時候就轉彎

一個簡單的dfs就可以啦

我這裡考慮的都是內部轉彎的情況 所以n<=3時都枚舉了一遍

程式碼如下：

#include<iostream>
#include<cstring>
using namespace std;
int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int map[40][40];
void print(int n)
{
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			if(map[i][j]) cout<<'#';
			else cout<<' ';
		}
		cout<<endl;
	}
}
void dfs(int n,int x,int y,int d){
	map[x][y]=1;
	int dn=(d+1)%4;
	int xn=x+dir[dn][0];
	int yn=y+dir[dn][1];
	int count=0;
	if(x+dir[d][0]<=n&&x+dir[d][0]>=1&&y+dir[d][1]>=1&&y+dir[d][1]<=n)
	{
		for(int i=0;i<4;i++)
		{
			if(map[xn+dir[i][0]][yn+dir[i][1]]) count++;
		}
		if(map[x+dir[d][0]*2][y+dir[d][1]*2]&&count>=2)
		{
			print(n);
			return ;
		}
	}
	if(x+dir[d][0]>n||x+dir[d][0]<1||y+dir[d][1]>n||y+dir[d][1]<1)
	{
		dfs(n,xn,yn,dn);
		return ;
	}
	else if(map[x+dir[d][0]*2][y+dir[d][1]*2])
	{
		dfs(n,xn,yn,dn);
		return ;
	}
	else
	{
		dfs(n,x+dir[d][0],y+dir[d][1],d);
	}
	return ;
	
}
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int n;
		cin>>n;
		memset(map,0,sizeof(map));
		if(n==1)
		{
			map[1][1]=1;
			print(n);
		}
		else if(n==2)
		{
			map[1][1]++;
			map[1][2]++;
			map[2][2]++;
			print(n);
		}
		else if(n==3)
		{
			map[1][1]++;
			map[1][2]++;
			map[1][3]++;
			map[2][3]++;
			map[3][3]++;
			map[3][2]++;
			map[3][1]++;
			print(n);
		}
		else
		{
			dfs(n,1,1,0);
		}
		
	}
}