#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, t, tr = 0;
  cin >> n;
  m = (n / 2) + 1;
  t = (n - 1);
  for (int i = 1; i <= n; i++) {
    if (i == m) {
      for (int j = 1; j <= n; j++) {
        cout << "D";
      }
      tr = 5;
      t = t + 2;
    } else if (tr == 5) {
      for (int k = 1; k <= t / 2; k++) {
        cout << "*";
      }
      for (int k = 1; k <= n - t; k++) cout << "D";
      for (int k = 1; k <= t / 2; k++) cout << "*";
      t = t + 2;
    } else {
      for (int k = 1; k <= t / 2; k++) {
        cout << "*";
      }
      for (int k = 1; k <= n - t; k++) cout << "D";
      for (int k = 1; k <= t / 2; k++) cout << "*";
      t = t - 2;
    }
    cout << endl;
  }
  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>
using namespace std;
int main() {
  int n, m, t, tr = 0;
  cin >> n;
  m = (n / 2) + 1;
  t = (n - 1);
  for (int i = 1; i <= n; i++) {
    if (i == m) {
      for (int j = 1; j <= n; j++) {
        cout << "D";
      }
      tr = 5;
      t = t + 2;
    } else if (tr == 5) {
      for (int k = 1; k <= t / 2; k++) {
        cout << "*";
      }
      for (int k = 1; k <= n - t; k++) cout << "D";
      for (int k = 1; k <= t / 2; k++) cout << "*";
      t = t + 2;
    } else {
      for (int k = 1; k <= t / 2; k++) {
        cout << "*";
      }
      for (int k = 1; k <= n - t; k++) cout << "D";
      for (int k = 1; k <= t / 2; k++) cout << "*";
      t = t - 2;
    }
    cout << endl;
  }
  return 0;
}

```