#include <bits/stdc++.h>
using namespace std;
const int MAX = 101;
const int inf = (1ll < 61);
int n, c;
char a[MAX][MAX];
int main() {
  cin >> n;
  memset(a, '*', sizeof(a));
  for (int i = 0; i < n / 2 + 1; i++) {
    for (int j = 0; j <= c; j++) {
      a[i][(n - 1) / 2 + j] = 'D';
      a[i][(n - 1) / 2 - j] = 'D';
      a[n - 1 - i][(n - 1) / 2 + j] = 'D';
      a[n - 1 - i][(n - 1) / 2 - j] = 'D';
    }
    c++;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) cout << a[i][j];
    cout << endl;
  }
  return 0;
}

### Prompt

Generate 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;
const int MAX = 101;
const int inf = (1ll < 61);
int n, c;
char a[MAX][MAX];
int main() {
  cin >> n;
  memset(a, '*', sizeof(a));
  for (int i = 0; i < n / 2 + 1; i++) {
    for (int j = 0; j <= c; j++) {
      a[i][(n - 1) / 2 + j] = 'D';
      a[i][(n - 1) / 2 - j] = 'D';
      a[n - 1 - i][(n - 1) / 2 + j] = 'D';
      a[n - 1 - i][(n - 1) / 2 - j] = 'D';
    }
    c++;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) cout << a[i][j];
    cout << endl;
  }
  return 0;
}

```