#include <bits/stdc++.h>
using namespace std;
int main() {
  int r, s, k = 0, i, d;
  cin >> r;
  d = (r / 2) + 1;
  for (i = 1; i <= d; i++, k = 0) {
    for (s = 1; s <= d - i; s++) {
      cout << "*";
    }
    while (k != 2 * i - 1) {
      cout << "D";
      k++;
    }
    for (s = 1; s <= d - i; s++) {
      cout << "*";
    }
    cout << endl;
  }
  for (i = d - 1; i >= 1; i--, k = 0) {
    for (s = d - i; s >= 1; s--) {
      cout << "*";
    }
    while (k != 2 * i - 1) {
      cout << "D";
      k++;
    }
    for (s = d - i; s >= 1; s--) {
      cout << "*";
    }
    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 r, s, k = 0, i, d;
  cin >> r;
  d = (r / 2) + 1;
  for (i = 1; i <= d; i++, k = 0) {
    for (s = 1; s <= d - i; s++) {
      cout << "*";
    }
    while (k != 2 * i - 1) {
      cout << "D";
      k++;
    }
    for (s = 1; s <= d - i; s++) {
      cout << "*";
    }
    cout << endl;
  }
  for (i = d - 1; i >= 1; i--, k = 0) {
    for (s = d - i; s >= 1; s--) {
      cout << "*";
    }
    while (k != 2 * i - 1) {
      cout << "D";
      k++;
    }
    for (s = d - i; s >= 1; s--) {
      cout << "*";
    }
    cout << endl;
  }
  return 0;
}

```