#include <bits/stdc++.h>
using namespace std;
long long int i, j, k;
string s;
long long int n, m, ans;
int main() {
  cin >> n;
  m = (n - 1) / 2;
  s = "";
  for (long long int i = 0; i < n; i++) {
    s += '*';
  }
  string s1[n];
  for (long long int i = 0; i < m + 1; i++) {
    s1[i] = s;
  }
  for (i = 0; i < (m + 1); i++) {
    for (j = (m - i); j <= (m + i); j++) {
      s1[i][j] = 'D';
    }
  }
  for (long long int i = 0; i < m; i++) {
    s1[n - 1 - i] = s1[i];
  }
  for (long long int i = 0; i < n; i++) {
    cout << s1[i] << 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;
long long int i, j, k;
string s;
long long int n, m, ans;
int main() {
  cin >> n;
  m = (n - 1) / 2;
  s = "";
  for (long long int i = 0; i < n; i++) {
    s += '*';
  }
  string s1[n];
  for (long long int i = 0; i < m + 1; i++) {
    s1[i] = s;
  }
  for (i = 0; i < (m + 1); i++) {
    for (j = (m - i); j <= (m + i); j++) {
      s1[i][j] = 'D';
    }
  }
  for (long long int i = 0; i < m; i++) {
    s1[n - 1 - i] = s1[i];
  }
  for (long long int i = 0; i < n; i++) {
    cout << s1[i] << endl;
  }
  return 0;
}

```