#include <bits/stdc++.h>
using namespace std;
template <typename T>
T power(T a, int n) {
  T res = 1;
  while (n) {
    if (n & 1) res = (res * a) % 1000000007;
    n >>= 1;
    a = (a * a) % 1000000007;
  }
  return res;
}
int main() {
  int n;
  int l, r, i;
  cin >> n;
  l = r = (n) / 2;
  i = 0;
  int j;
  while (i <= n / 2) {
    j = 0;
    while (j < n) {
      cout << (l <= j && j <= r ? 'D' : '*');
      j++;
    }
    printf("\n");
    ;
    l--;
    r++;
    i++;
  }
  l = 1;
  r = n - 2;
  while (i < n) {
    j = 0;
    while (j < n) {
      cout << (l <= j && j <= r ? 'D' : '*');
      j++;
    }
    printf("\n");
    ;
    l++;
    r--;
    i++;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
template <typename T>
T power(T a, int n) {
  T res = 1;
  while (n) {
    if (n & 1) res = (res * a) % 1000000007;
    n >>= 1;
    a = (a * a) % 1000000007;
  }
  return res;
}
int main() {
  int n;
  int l, r, i;
  cin >> n;
  l = r = (n) / 2;
  i = 0;
  int j;
  while (i <= n / 2) {
    j = 0;
    while (j < n) {
      cout << (l <= j && j <= r ? 'D' : '*');
      j++;
    }
    printf("\n");
    ;
    l--;
    r++;
    i++;
  }
  l = 1;
  r = n - 2;
  while (i < n) {
    j = 0;
    while (j < n) {
      cout << (l <= j && j <= r ? 'D' : '*');
      j++;
    }
    printf("\n");
    ;
    l++;
    r--;
    i++;
  }
  return 0;
}

```