#include <bits/stdc++.h>
using namespace std;
void filein() { freopen("input.txt", "r", stdin); }
void fileout() { freopen("output.txt", "w", stdout); }
long long n, a[10000], sm = 0;
int main() {
  cin >> n;
  int h = 1;
  int j1 = n / 2;
  for (int i = 1; i <= n / 2; i++) {
    for (int j = 1; j <= j1 - i + 1; j++) cout << "*";
    for (int hh = 1; hh <= h; hh++) cout << "D";
    h += 2;
    for (int j = 1; j <= j1 - i + 1; j++) cout << "*";
    cout << endl;
  }
  for (int i = 0; i < n; i++) cout << "D";
  cout << endl;
  h = n - 2;
  j1 = n / 2;
  for (int i = 1; i <= n / 2; i++) {
    for (int j = 1; j <= (n - h) / 2; j++) cout << "*";
    for (int hh = 1; hh <= h; hh++) cout << "D";
    for (int j = 1; j <= (n - h) / 2; j++) cout << "*";
    h -= 2;
    cout << endl;
  }
}

### Prompt

Please create a solution in CPP 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;
void filein() { freopen("input.txt", "r", stdin); }
void fileout() { freopen("output.txt", "w", stdout); }
long long n, a[10000], sm = 0;
int main() {
  cin >> n;
  int h = 1;
  int j1 = n / 2;
  for (int i = 1; i <= n / 2; i++) {
    for (int j = 1; j <= j1 - i + 1; j++) cout << "*";
    for (int hh = 1; hh <= h; hh++) cout << "D";
    h += 2;
    for (int j = 1; j <= j1 - i + 1; j++) cout << "*";
    cout << endl;
  }
  for (int i = 0; i < n; i++) cout << "D";
  cout << endl;
  h = n - 2;
  j1 = n / 2;
  for (int i = 1; i <= n / 2; i++) {
    for (int j = 1; j <= (n - h) / 2; j++) cout << "*";
    for (int hh = 1; hh <= h; hh++) cout << "D";
    for (int j = 1; j <= (n - h) / 2; j++) cout << "*";
    h -= 2;
    cout << endl;
  }
}

```