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

### Prompt

Your task is to create 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 n;
  cin >> n;
  char a[n][n];
  for (int i = 0; i < n; i++) {
    if (i <= ((n - 1) / 2)) {
      for (int j = 0; j < n; j++) {
        if (((n - 1) / 2 - i <= j) && (j <= (n - 1) / 2 + i)) {
          a[i][j] = 'D';
        } else {
          a[i][j] = '*';
        }
      }
    } else {
      for (int l = 0; l < n; l++) {
        a[i][l] = a[n - 1 - i][l];
      }
    }
  }
  for (int l = 0; l < n; l++) {
    for (int m = 0; m < n; m++) {
      cout << a[l][m];
    }
    cout << endl;
  }
  return 0;
}

```