#include <bits/stdc++.h>
using namespace std;
map<string, long long> mp, pm;
vector<long long> v, p;
long long a, n, d, c;
string g, h;
char t, y, u;
int main() {
  cout.tie(NULL);
  cin.tie(NULL);
  ios_base::sync_with_stdio(0);
  cin >> a;
  d = 1;
  n = a / 2;
  c = a / 2 + 1;
  for (int i = 1; i <= a; i++) {
    if (c == i) {
      d -= 2;
      n = 1;
      for (int j = 0; j < a; j++) {
        cout << "D";
      }
      cout << endl;
    }
    if (c > i) {
      for (int j = 0; j < n; j++) {
        cout << "*";
      }
      for (int j = 0; j < d; j++) {
        cout << "D";
      }
      for (int j = 0; j < n; j++) cout << "*";
      cout << endl;
      n--;
      d += 2;
    } else if (c < i) {
      for (int j = 0; j < n; j++) cout << "*";
      for (int j = 0; j < d; j++) cout << "D";
      for (int j = 0; j < n; j++) cout << "*";
      cout << endl;
      d -= 2;
      n++;
    }
  }
}

### 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;
map<string, long long> mp, pm;
vector<long long> v, p;
long long a, n, d, c;
string g, h;
char t, y, u;
int main() {
  cout.tie(NULL);
  cin.tie(NULL);
  ios_base::sync_with_stdio(0);
  cin >> a;
  d = 1;
  n = a / 2;
  c = a / 2 + 1;
  for (int i = 1; i <= a; i++) {
    if (c == i) {
      d -= 2;
      n = 1;
      for (int j = 0; j < a; j++) {
        cout << "D";
      }
      cout << endl;
    }
    if (c > i) {
      for (int j = 0; j < n; j++) {
        cout << "*";
      }
      for (int j = 0; j < d; j++) {
        cout << "D";
      }
      for (int j = 0; j < n; j++) cout << "*";
      cout << endl;
      n--;
      d += 2;
    } else if (c < i) {
      for (int j = 0; j < n; j++) cout << "*";
      for (int j = 0; j < d; j++) cout << "D";
      for (int j = 0; j < n; j++) cout << "*";
      cout << endl;
      d -= 2;
      n++;
    }
  }
}

```