#include <bits/stdc++.h>
using namespace std;
int n;
inline void star(const int& n, const int& i) {
  for (int j = n; j > i; --j) {
    cout << '*';
  }
  return;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    if (i + 1 == n / 2 + 1) {
      for (int j = 0; j < n; ++j) {
        cout << 'D';
      }
    } else if (i < n / 2) {
      star(n / 2, i);
      for (int j = 2 * i + 1; j > 0; --j) {
        cout << 'D';
      }
      star(n / 2, i);
    } else {
      star(i, n / 2);
      cout << 'D';
      if (i != n - 1) {
        for (int j = i + 1; j < n; ++j) {
          cout << "DD";
        }
      }
      star(i, n / 2);
    }
    cout << endl;
  }
  return false;
}

### 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>
using namespace std;
int n;
inline void star(const int& n, const int& i) {
  for (int j = n; j > i; --j) {
    cout << '*';
  }
  return;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    if (i + 1 == n / 2 + 1) {
      for (int j = 0; j < n; ++j) {
        cout << 'D';
      }
    } else if (i < n / 2) {
      star(n / 2, i);
      for (int j = 2 * i + 1; j > 0; --j) {
        cout << 'D';
      }
      star(n / 2, i);
    } else {
      star(i, n / 2);
      cout << 'D';
      if (i != n - 1) {
        for (int j = i + 1; j < n; ++j) {
          cout << "DD";
        }
      }
      star(i, n / 2);
    }
    cout << endl;
  }
  return false;
}

```