#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  while (cin >> n) {
    int i = 0, y = 1;
    for (; y <= n / 2 + 1; y++) {
      for (int x = 1; x <= n; x++) {
        if (x < n / 2 + 1 - i || x > n / 2 + 1 + i)
          cout << '*';
        else
          cout << 'D';
      }
      cout << "\n";
      i++;
    }
    i = n / 2 - 1;
    for (; y <= n; y++) {
      for (int x = 1; x <= n; x++) {
        if (x < n / 2 + 1 - i || x > n / 2 + 1 + i)
          cout << '*';
        else
          cout << 'D';
      }
      cout << "\n";
      --i;
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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>
using namespace std;
int main() {
  int n;
  while (cin >> n) {
    int i = 0, y = 1;
    for (; y <= n / 2 + 1; y++) {
      for (int x = 1; x <= n; x++) {
        if (x < n / 2 + 1 - i || x > n / 2 + 1 + i)
          cout << '*';
        else
          cout << 'D';
      }
      cout << "\n";
      i++;
    }
    i = n / 2 - 1;
    for (; y <= n; y++) {
      for (int x = 1; x <= n; x++) {
        if (x < n / 2 + 1 - i || x > n / 2 + 1 + i)
          cout << '*';
        else
          cout << 'D';
      }
      cout << "\n";
      --i;
    }
    cout << endl;
  }
  return 0;
}

```