#include <bits/stdc++.h>
using namespace std;
const long long max1 = 101;
int main() {
  char a[51][101];
  bool debug = false;
  int otkl = 0;
  if (debug) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  int n;
  cin >> n;
  for (size_t i = 0; i < (n + 1) / 2; i++) {
    for (size_t j = 0; j < n; j++) {
      if ((j <= ((n - 1) / 2) + otkl) && (j >= ((n - 1) / 2) - otkl)) {
        a[i][j] = 'D';
        cout << a[i][j];
      } else {
        a[i][j] = '*';
        cout << a[i][j];
      }
    }
    cout << endl;
    otkl++;
  }
  long long nas = ((n - 1) / 2) - 1;
  for (long long i = nas; i >= 0; i--) {
    for (size_t j = 0; j < n; j++) {
      cout << a[i][j];
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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 long long max1 = 101;
int main() {
  char a[51][101];
  bool debug = false;
  int otkl = 0;
  if (debug) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  int n;
  cin >> n;
  for (size_t i = 0; i < (n + 1) / 2; i++) {
    for (size_t j = 0; j < n; j++) {
      if ((j <= ((n - 1) / 2) + otkl) && (j >= ((n - 1) / 2) - otkl)) {
        a[i][j] = 'D';
        cout << a[i][j];
      } else {
        a[i][j] = '*';
        cout << a[i][j];
      }
    }
    cout << endl;
    otkl++;
  }
  long long nas = ((n - 1) / 2) - 1;
  for (long long i = nas; i >= 0; i--) {
    for (size_t j = 0; j < n; j++) {
      cout << a[i][j];
    }
    cout << endl;
  }
  return 0;
}

```