#include <bits/stdc++.h>
using namespace std;
string a[101], s1, s;
int main() {
  int n;
  cin >> n;
  int mid = n / 2 + 1;
  for (int i = 1; i <= n / 2; i++) {
    s1 = "", s = "";
    for (int j = 1; j <= mid - i; j++) s1 = s1 + "*";
    for (int j = 1; j <= 2 * i - 1; j++) s = s + "D";
    s = s1 + s + s1;
    a[i] = s;
  }
  for (int i = 0; i < n; i++) a[mid] += "D";
  for (int i = mid + 1; i <= n; i++) {
    s = "", s1 = "";
    for (int j = 1; j <= i - mid; j++) s1 = s1 + "*";
    for (int j = 1; j <= n - 2 * (i - mid); j++) s = s + "D";
    a[i] = s1 + s + s1;
  }
  for (int i = 1; i <= n; i++) cout << a[i] << "\n";
}

### 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;
string a[101], s1, s;
int main() {
  int n;
  cin >> n;
  int mid = n / 2 + 1;
  for (int i = 1; i <= n / 2; i++) {
    s1 = "", s = "";
    for (int j = 1; j <= mid - i; j++) s1 = s1 + "*";
    for (int j = 1; j <= 2 * i - 1; j++) s = s + "D";
    s = s1 + s + s1;
    a[i] = s;
  }
  for (int i = 0; i < n; i++) a[mid] += "D";
  for (int i = mid + 1; i <= n; i++) {
    s = "", s1 = "";
    for (int j = 1; j <= i - mid; j++) s1 = s1 + "*";
    for (int j = 1; j <= n - 2 * (i - mid); j++) s = s + "D";
    a[i] = s1 + s + s1;
  }
  for (int i = 1; i <= n; i++) cout << a[i] << "\n";
}

```