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

```