#include <bits/stdc++.h>
const int INF = 10e9 + 7;
const int MXN = 10e5 + 1;
using namespace std;
long long n, l, r;
char ch[110][110];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      ch[i][j] = '*';
    }
  }
  l = (n + 1) / 2;
  r = l;
  while (l > 1 && r < n) {
    for (int i = 1; i <= (n + 1) / 2; i++) {
      for (int j = l; j <= r; j++) {
        ch[i][j] = 'D';
      }
      l--;
      r++;
    }
  }
  l = 2;
  r = n - 1;
  while (l <= r) {
    for (int i = (n + 1) / 2 + 1; i <= n; i++) {
      for (int j = l; j <= r; j++) {
        ch[i][j] = 'D';
      }
      l++;
      r--;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cout << ch[i][j];
    }
    cout << endl;
  }
  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>
const int INF = 10e9 + 7;
const int MXN = 10e5 + 1;
using namespace std;
long long n, l, r;
char ch[110][110];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      ch[i][j] = '*';
    }
  }
  l = (n + 1) / 2;
  r = l;
  while (l > 1 && r < n) {
    for (int i = 1; i <= (n + 1) / 2; i++) {
      for (int j = l; j <= r; j++) {
        ch[i][j] = 'D';
      }
      l--;
      r++;
    }
  }
  l = 2;
  r = n - 1;
  while (l <= r) {
    for (int i = (n + 1) / 2 + 1; i <= n; i++) {
      for (int j = l; j <= r; j++) {
        ch[i][j] = 'D';
      }
      l++;
      r--;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cout << ch[i][j];
    }
    cout << endl;
  }
  return 0;
}

```