#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  string s;
  for (int i = 0; i < n; ++i) {
    s.push_back('*');
  }
  vector<string> grid(n, s);
  int start = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < start; j++) {
      grid[i][n / 2 + j] = 'D';
      grid[i][n / 2 - j] = 'D';
    }
    if (i < n / 2) {
      ++start;
    } else {
      --start;
    }
  }
  for (string s : grid) {
    cout << s << endl;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
  cin >> n;
  string s;
  for (int i = 0; i < n; ++i) {
    s.push_back('*');
  }
  vector<string> grid(n, s);
  int start = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < start; j++) {
      grid[i][n / 2 + j] = 'D';
      grid[i][n / 2 - j] = 'D';
    }
    if (i < n / 2) {
      ++start;
    } else {
      --start;
    }
  }
  for (string s : grid) {
    cout << s << endl;
  }
  return 0;
}

```