#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int k = (n / 2) + 1;
  int t = 1;
  char a[105][105];
  int x = 0, y = 0;
  for (int i = 1; i <= n / 2; i++) {
    int c = 0;
    for (int j = 1; j <= n; j++) {
      if (c == n) break;
      if (k == j)
        for (int l = 0; l < t; l++) {
          c++;
          cout << "D";
        }
      else {
        c++;
        cout << "*";
      }
    }
    cout << endl;
    t += 2;
    k--;
  }
  for (int i = 1; i <= n; i++) cout << "D";
  cout << endl;
  k = 2;
  t = n - 2;
  for (int i = 1; i <= n / 2; i++) {
    int c = 0;
    for (int j = 1; j <= n; j++) {
      if (c == n) break;
      if (k == j)
        for (int l = 0; l < t; l++) {
          c++;
          cout << "D";
        }
      else {
        c++;
        cout << "*";
      }
    }
    cout << endl;
    t -= 2;
    k++;
  }
  return 0;
}

### Prompt

Your challenge is to write 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() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int k = (n / 2) + 1;
  int t = 1;
  char a[105][105];
  int x = 0, y = 0;
  for (int i = 1; i <= n / 2; i++) {
    int c = 0;
    for (int j = 1; j <= n; j++) {
      if (c == n) break;
      if (k == j)
        for (int l = 0; l < t; l++) {
          c++;
          cout << "D";
        }
      else {
        c++;
        cout << "*";
      }
    }
    cout << endl;
    t += 2;
    k--;
  }
  for (int i = 1; i <= n; i++) cout << "D";
  cout << endl;
  k = 2;
  t = n - 2;
  for (int i = 1; i <= n / 2; i++) {
    int c = 0;
    for (int j = 1; j <= n; j++) {
      if (c == n) break;
      if (k == j)
        for (int l = 0; l < t; l++) {
          c++;
          cout << "D";
        }
      else {
        c++;
        cout << "*";
      }
    }
    cout << endl;
    t -= 2;
    k++;
  }
  return 0;
}

```