#include <bits/stdc++.h>
int main() {
  int mid, cnt = -1, i, j, n;
  scanf("%d", &n);
  char* arr = (char*)malloc((n * n) * sizeof(char));
  memset(arr, 42, (n * n) * sizeof(char));
  mid = n / 2;
  for (j = 0; j < n; j++) {
    if (j <= n / 2)
      cnt++;
    else
      cnt--;
    for (i = mid - cnt; i <= mid + cnt; i++) arr[n * j + i] = 'D';
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      printf("%c", arr[n * i + j]);
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n × n matrix with a diamond inscribed into it.

You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.

Input

The only line contains an integer n (3 ≤ n ≤ 101; n is odd). 

Output

Output a crystal of size n.

Examples

Input

3


Output

*D*
DDD
*D*


Input

5


Output

**D**
*DDD*
DDDDD
*DDD*
**D**


Input

7


Output

***D***
**DDD**
*DDDDD*
DDDDDDD
*DDDDD*
**DDD**
***D***

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  int mid, cnt = -1, i, j, n;
  scanf("%d", &n);
  char* arr = (char*)malloc((n * n) * sizeof(char));
  memset(arr, 42, (n * n) * sizeof(char));
  mid = n / 2;
  for (j = 0; j < n; j++) {
    if (j <= n / 2)
      cnt++;
    else
      cnt--;
    for (i = mid - cnt; i <= mid + cnt; i++) arr[n * j + i] = 'D';
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      printf("%c", arr[n * i + j]);
    }
    printf("\n");
  }
  return 0;
}

```