#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j, cnt = 0, x, k;
  cin >> n;
  x = n / 2 + 1;
  for (i = 1; i <= n / 2; i++) {
    for (j = 1; j <= n; j++) {
      if (j >= x && j <= x + cnt)
        cout << "D";
      else if (j >= n / 2 + 1 && j <= n / 2 + 1 + cnt)
        cout << "D";
      else
        cout << "*";
    }
    x--;
    cnt++;
    cout << "\n";
  }
  for (i = 0; i < n; i++) cout << "D";
  cout << "\n";
  cnt = 1;
  for (i = 1; i <= n / 2; i++) {
    for (j = 1; j <= cnt; j++) cout << "*";
    for (j = cnt + 1; j <= n - cnt; j++) cout << "D";
    for (j = n - cnt + 1; j <= n; j++) cout << "*";
    cnt++;
    cout << "\n";
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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, i, j, cnt = 0, x, k;
  cin >> n;
  x = n / 2 + 1;
  for (i = 1; i <= n / 2; i++) {
    for (j = 1; j <= n; j++) {
      if (j >= x && j <= x + cnt)
        cout << "D";
      else if (j >= n / 2 + 1 && j <= n / 2 + 1 + cnt)
        cout << "D";
      else
        cout << "*";
    }
    x--;
    cnt++;
    cout << "\n";
  }
  for (i = 0; i < n; i++) cout << "D";
  cout << "\n";
  cnt = 1;
  for (i = 1; i <= n / 2; i++) {
    for (j = 1; j <= cnt; j++) cout << "*";
    for (j = cnt + 1; j <= n - cnt; j++) cout << "D";
    for (j = n - cnt + 1; j <= n; j++) cout << "*";
    cnt++;
    cout << "\n";
  }
}

```