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

### Prompt

Please formulate 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 i, j, n, m, k, s;
  cin >> n;
  m = (n / 2) + 1;
  s = 1;
  for (i = 0; i < m; i++) {
    for (j = m - 1; j > i; j--) {
      cout << "*";
    }
    for (j = 1; j <= s; j++) {
      cout << "D";
    }
    for (j = m - 1; j > i; j--) {
      cout << "*";
    }
    s = s + 2;
    cout << "\n";
  }
  s = n - 2;
  for (i = 0; i < (m - 1); i++) {
    for (j = 0; j <= i; j++) {
      cout << "*";
    }
    for (j = 1; j <= s; j++) {
      cout << "D";
    }
    for (j = 0; j <= i; j++) {
      cout << "*";
    }
    s = s - 2;
    if (i != (m - 2)) {
      cout << "\n";
    }
  }
  return 0;
}

```