#include<iostream>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<string>
#include<cstring>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<bitset>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> i_i;
typedef pair<long long, int> ll_i;

#define PI 3.141592653589793238462643383279
#define mod 1000000007LL
#define rep(i, n) for(i = 0;i < n;++i)
#define rep1(i, n) for(i = 1;i < n;++i)
#define per(i, n) for(i = n - 1;i > -1;--i)
#define int(x) int x; scanf("%d",&x)
#define int2(x, y) int x, y; scanf("%d%d",&x, &y)
#define int3(x, y, z) int x, y, z; scanf("%d%d%d",&x, &y, &z)
#define int4(v, x, y, z) int v, x, y, z; scanf("%d%d%d%d", &v, &x, &y, &z)
#define int5(v, w, x, y, z) int v, w, x, y, z; scanf("%d%d%d%d%d", &v, &w, &x, &y, &z)
#define pri(x) cout << (x) << "\n"
#define pri2(x, y) cout << (x) << " " << (y) << "\n"
#define pri3(x, y, z) cout << (x) << " " << (y) << " " << (z) << "\n"
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(),(a).end()
#define kabe puts("---------------------------")
#define kara puts("")
#define debug(x) cout << " --- " << (x) << "\n"
#define debug2(x, y) cout << " --- " << (x) << " " << (y) << "\n"
#define debug3(x, y, z) cout << " --- " << (x) << " " << (y) << " " << (z) << "\n"
#define X first
#define Y second
#define eps 0.0001
#define prid(x) printf("%.15lf\n", x)

char f[100][100];
int dw[4] = {0, 1, 0, -1};
int dh[4] = {-1, 0, 1, 0};
signed main(void){
  int i, j;
  int(d);
  for(int testcase = 0;testcase < d;testcase++){
    if(testcase != 0)puts("");
    int(n);
    rep(i, n)rep(j, n)f[i][j] = ' ';
    int noww = 0, nowh = n - 1, k = 0;

    bool is = false;
    for(;;){
      //printf("%d %d %d\n", nowh, noww, k);
      f[nowh][noww] = '#';
      if(0 > nowh + dh[k] || nowh + dh[k] >= n || 0 > noww + dw[k] || noww + dw[k] >= n){
        //puts("a");
        if(is)break;
        is = true;
        k = (k + 1) % 4;
        continue;
      }
      if(0 <= nowh + dh[k] * 2 && nowh + dh[k] * 2 < n)
      if(0 <= noww + dw[k] * 2 && noww + dw[k] * 2 < n)
      if(f[nowh + dh[k] + dh[k]][noww + dw[k] + dw[k]] == '#'){
        //puts("b");
        if(is)break;
        is = true;
        k = (k + 1) % 4;
        continue;
      }
      if(0 <= nowh + dh[k] + dh[(k + 1) % 4] && nowh + dh[k] + dh[(k + 1) % 4] < n)
      if(0 <= noww + dw[k] + dw[(k + 1) % 4] && noww + dw[k] + dw[(k + 1) % 4] < n)
      if(f[nowh + dh[k] + dh[(k + 1) % 4]][noww + dw[k] + dw[(k + 1) % 4]] == '#'){
        //puts("c");
        if(is)break;
        is = true;
        k = (k + 1) % 4;
        continue;
      }
      if(0 <= nowh + dh[(k + 3) % 4] && nowh + dh[(k + 3) % 4] < n)
      if(0 <= noww + dw[(k + 3) % 4] && noww + dw[(k + 3) % 4] < n)
      if(f[nowh + dh[(k + 3) % 4]][noww + dw[(k + 3) % 4]] == '#'){
        //puts("d");
        if(is)break;
        is = true;
        k = (k + 1) % 4;
        continue;
      }
      //puts("e");
      is = false;
      nowh += dh[k]; noww += dw[k];
    }

    rep(i, n){
      rep(j, n)printf("%c", f[i][j]);
      puts("");
    }

//*/

//*/ break;
  }
  return 0;
}
### Prompt

Generate 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<algorithm>
#include<climits>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<string>
#include<cstring>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<bitset>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> i_i;
typedef pair<long long, int> ll_i;

#define PI 3.141592653589793238462643383279
#define mod 1000000007LL
#define rep(i, n) for(i = 0;i < n;++i)
#define rep1(i, n) for(i = 1;i < n;++i)
#define per(i, n) for(i = n - 1;i > -1;--i)
#define int(x) int x; scanf("%d",&x)
#define int2(x, y) int x, y; scanf("%d%d",&x, &y)
#define int3(x, y, z) int x, y, z; scanf("%d%d%d",&x, &y, &z)
#define int4(v, x, y, z) int v, x, y, z; scanf("%d%d%d%d", &v, &x, &y, &z)
#define int5(v, w, x, y, z) int v, w, x, y, z; scanf("%d%d%d%d%d", &v, &w, &x, &y, &z)
#define pri(x) cout << (x) << "\n"
#define pri2(x, y) cout << (x) << " " << (y) << "\n"
#define pri3(x, y, z) cout << (x) << " " << (y) << " " << (z) << "\n"
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(),(a).end()
#define kabe puts("---------------------------")
#define kara puts("")
#define debug(x) cout << " --- " << (x) << "\n"
#define debug2(x, y) cout << " --- " << (x) << " " << (y) << "\n"
#define debug3(x, y, z) cout << " --- " << (x) << " " << (y) << " " << (z) << "\n"
#define X first
#define Y second
#define eps 0.0001
#define prid(x) printf("%.15lf\n", x)

char f[100][100];
int dw[4] = {0, 1, 0, -1};
int dh[4] = {-1, 0, 1, 0};
signed main(void){
  int i, j;
  int(d);
  for(int testcase = 0;testcase < d;testcase++){
    if(testcase != 0)puts("");
    int(n);
    rep(i, n)rep(j, n)f[i][j] = ' ';
    int noww = 0, nowh = n - 1, k = 0;

    bool is = false;
    for(;;){
      //printf("%d %d %d\n", nowh, noww, k);
      f[nowh][noww] = '#';
      if(0 > nowh + dh[k] || nowh + dh[k] >= n || 0 > noww + dw[k] || noww + dw[k] >= n){
        //puts("a");
        if(is)break;
        is = true;
        k = (k + 1) % 4;
        continue;
      }
      if(0 <= nowh + dh[k] * 2 && nowh + dh[k] * 2 < n)
      if(0 <= noww + dw[k] * 2 && noww + dw[k] * 2 < n)
      if(f[nowh + dh[k] + dh[k]][noww + dw[k] + dw[k]] == '#'){
        //puts("b");
        if(is)break;
        is = true;
        k = (k + 1) % 4;
        continue;
      }
      if(0 <= nowh + dh[k] + dh[(k + 1) % 4] && nowh + dh[k] + dh[(k + 1) % 4] < n)
      if(0 <= noww + dw[k] + dw[(k + 1) % 4] && noww + dw[k] + dw[(k + 1) % 4] < n)
      if(f[nowh + dh[k] + dh[(k + 1) % 4]][noww + dw[k] + dw[(k + 1) % 4]] == '#'){
        //puts("c");
        if(is)break;
        is = true;
        k = (k + 1) % 4;
        continue;
      }
      if(0 <= nowh + dh[(k + 3) % 4] && nowh + dh[(k + 3) % 4] < n)
      if(0 <= noww + dw[(k + 3) % 4] && noww + dw[(k + 3) % 4] < n)
      if(f[nowh + dh[(k + 3) % 4]][noww + dw[(k + 3) % 4]] == '#'){
        //puts("d");
        if(is)break;
        is = true;
        k = (k + 1) % 4;
        continue;
      }
      //puts("e");
      is = false;
      nowh += dh[k]; noww += dw[k];
    }

    rep(i, n){
      rep(j, n)printf("%c", f[i][j]);
      puts("");
    }

//*/

//*/ break;
  }
  return 0;
}
```