#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, t, m = 0, k, f;
  cin >> n;
  t = (((n + 1) / 2) - 1);
  f = (n + 1) / 2;
  char a[101][101];
  for (int i = 0; i < f; i++) {
    a[i][t] = 'D';
    k = 1;
    for (int j = 0; j < m; j++) {
      a[i][t + k] = 'D';
      a[i][t - k] = 'D';
      k++;
    }
    m++;
  }
  m = ((n - 3) / 2);
  for (int i = f; i < n; i++) {
    a[i][t] = 'D';
    k = 1;
    for (int j = 0; j < m; j++) {
      a[i][t + k] = 'D';
      a[i][t - k] = 'D';
      k++;
    }
    m--;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (a[i - 1][j] != 'D' && a[i + 1][j] != 'D' && a[i][j + 1] != 'D' &&
          a[i][j - 1] != 'D') {
        a[i][j] = '*';
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (a[i][j] != 'D') {
        a[i][j] = '*';
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << a[i][j];
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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() {
  long long n, t, m = 0, k, f;
  cin >> n;
  t = (((n + 1) / 2) - 1);
  f = (n + 1) / 2;
  char a[101][101];
  for (int i = 0; i < f; i++) {
    a[i][t] = 'D';
    k = 1;
    for (int j = 0; j < m; j++) {
      a[i][t + k] = 'D';
      a[i][t - k] = 'D';
      k++;
    }
    m++;
  }
  m = ((n - 3) / 2);
  for (int i = f; i < n; i++) {
    a[i][t] = 'D';
    k = 1;
    for (int j = 0; j < m; j++) {
      a[i][t + k] = 'D';
      a[i][t - k] = 'D';
      k++;
    }
    m--;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (a[i - 1][j] != 'D' && a[i + 1][j] != 'D' && a[i][j + 1] != 'D' &&
          a[i][j - 1] != 'D') {
        a[i][j] = '*';
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (a[i][j] != 'D') {
        a[i][j] = '*';
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << a[i][j];
    }
    cout << endl;
  }
  return 0;
}

```