#include <bits/stdc++.h>
using namespace std;
int n;
void print(int x) {
  int ast = (n - x) / 2;
  for (int i = 0; i < ast; i++) {
    cout << "*";
  }
  for (int i = 0; i < x; i++) {
    cout << "D";
  }
  for (int i = 0; i < ast; i++) {
    cout << "*";
  }
  cout << endl;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin >> n;
  int d = 1;
  while (d < n) {
    print(d);
    d += 2;
  }
  print(d);
  d -= 2;
  while (d > 0) {
    print(d);
    d -= 2;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
void print(int x) {
  int ast = (n - x) / 2;
  for (int i = 0; i < ast; i++) {
    cout << "*";
  }
  for (int i = 0; i < x; i++) {
    cout << "D";
  }
  for (int i = 0; i < ast; i++) {
    cout << "*";
  }
  cout << endl;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin >> n;
  int d = 1;
  while (d < n) {
    print(d);
    d += 2;
  }
  print(d);
  d -= 2;
  while (d > 0) {
    print(d);
    d -= 2;
  }
  return 0;
}

```