#include <bits/stdc++.h>
bool a[111][111];
int main() {
  int n, m, i, j;
  scanf("%d", &n);
  memset(a, false, sizeof(a));
  m = n / 2;
  for (i = 0; i <= m; ++i) {
    for (j = 0; j <= i; ++j) {
      a[i][m - j] = a[i][m + j] = true;
      a[n - i - 1][m - j] = a[n - i - 1][m + j] = true;
    }
  }
  for (i = 0; i < n; ++i) {
    for (j = 0; j < n; ++j) printf("%c", a[i][j] ? 'D' : '*');
    printf("\n");
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
bool a[111][111];
int main() {
  int n, m, i, j;
  scanf("%d", &n);
  memset(a, false, sizeof(a));
  m = n / 2;
  for (i = 0; i <= m; ++i) {
    for (j = 0; j <= i; ++j) {
      a[i][m - j] = a[i][m + j] = true;
      a[n - i - 1][m - j] = a[n - i - 1][m + j] = true;
    }
  }
  for (i = 0; i < n; ++i) {
    for (j = 0; j < n; ++j) printf("%c", a[i][j] ? 'D' : '*');
    printf("\n");
  }
  return 0;
}

```