#include <bits/stdc++.h>
using namespace std;
int n, x;
char t[200][200];
string ch, ch1;
bool b = 0;
int main() {
  cin >> n;
  int cur = n / 2;
  int cul = n / 2;
  for (int i = 0; i < n; i++) {
    for (int j = cur; j <= cul; j++) {
      t[i][j] = 'D';
    }
    if (cur == 0 && cul == n - 1) b = 1;
    if (b) {
      cur++;
      cul--;
    } else {
      cul++;
      cur--;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (t[i][j] != 'D')
        cout << '*';
      else
        cout << t[i][j];
    }
    cout << endl;
  }
}

### 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;
int n, x;
char t[200][200];
string ch, ch1;
bool b = 0;
int main() {
  cin >> n;
  int cur = n / 2;
  int cul = n / 2;
  for (int i = 0; i < n; i++) {
    for (int j = cur; j <= cul; j++) {
      t[i][j] = 'D';
    }
    if (cur == 0 && cul == n - 1) b = 1;
    if (b) {
      cur++;
      cul--;
    } else {
      cul++;
      cur--;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (t[i][j] != 'D')
        cout << '*';
      else
        cout << t[i][j];
    }
    cout << endl;
  }
}

```