#include <bits/stdc++.h>
using namespace std;
char a[200][200];
int main() {
  int n, k = 0;
  cin >> n;
  int m;
  char c = 'D';
  m = ((n - 1) / 2) + 1;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      if (j == ((n + 1) / 2) - 1) {
        a[i][j] = c;
      } else if (j != ((n + 1) / 2) - 1) {
        a[i][j] = '*';
      }
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      if (a[i - 1][j + 1] == c) {
        a[i][j] = c;
        a[i][j + 1] = c;
      }
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = n - 1; j >= 0; j--) {
      if (a[i - 1][j - 1] == c) {
        a[i][j] = c;
        a[i][j - 1] = c;
      }
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) cout << a[i][j];
    cout << endl;
  }
  for (int i = m - 2; i >= 0; i--) {
    for (int j = 0; j < n; j++) cout << a[i][j];
    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;
char a[200][200];
int main() {
  int n, k = 0;
  cin >> n;
  int m;
  char c = 'D';
  m = ((n - 1) / 2) + 1;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      if (j == ((n + 1) / 2) - 1) {
        a[i][j] = c;
      } else if (j != ((n + 1) / 2) - 1) {
        a[i][j] = '*';
      }
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      if (a[i - 1][j + 1] == c) {
        a[i][j] = c;
        a[i][j + 1] = c;
      }
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = n - 1; j >= 0; j--) {
      if (a[i - 1][j - 1] == c) {
        a[i][j] = c;
        a[i][j - 1] = c;
      }
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) cout << a[i][j];
    cout << endl;
  }
  for (int i = m - 2; i >= 0; i--) {
    for (int j = 0; j < n; j++) cout << a[i][j];
    cout << endl;
  }
  return 0;
}

```