#include <bits/stdc++.h>
using namespace std;
int main() {
  int a;
  cin >> a;
  int r = a / 2;
  for (int i = 0; i < a / 2; i++) {
    for (int y = 0; y < r; y++) {
      cout << "*";
    }
    for (int p = 0; p < a - r * 2; p++) {
      cout << "D";
    }
    for (int o = 0; o < r; o++) {
      cout << "*";
    }
    cout << endl;
    r--;
  }
  for (int i = 0; i < a / 2 + 1; i++) {
    for (int y = 0; y < r; y++) {
      cout << "*";
    }
    for (int p = 0; p < a - r * 2; p++) {
      cout << "D";
    }
    for (int o = 0; o < r; o++) {
      cout << "*";
    }
    cout << endl;
    r++;
  }
  return 0;
}

### 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 a;
  cin >> a;
  int r = a / 2;
  for (int i = 0; i < a / 2; i++) {
    for (int y = 0; y < r; y++) {
      cout << "*";
    }
    for (int p = 0; p < a - r * 2; p++) {
      cout << "D";
    }
    for (int o = 0; o < r; o++) {
      cout << "*";
    }
    cout << endl;
    r--;
  }
  for (int i = 0; i < a / 2 + 1; i++) {
    for (int y = 0; y < r; y++) {
      cout << "*";
    }
    for (int p = 0; p < a - r * 2; p++) {
      cout << "D";
    }
    for (int o = 0; o < r; o++) {
      cout << "*";
    }
    cout << endl;
    r++;
  }
  return 0;
}

```