#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void smax(T &x, T y) {
  x = max((x), (y));
}
template <class T>
inline void smin(T &x, T y) {
  x = min((x), (y));
}
inline void scanInt(int &x) {
  register char c = getchar();
  x = 0;
  bool neg = false;
  while (c < 48 || c > 57) {
    if (c == '-') neg = true;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x *= -1;
}
int main(int argc, const char *argv[]) {
  int N;
  scanInt(N);
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if ((abs(i - N / 2) + abs(N / 2 - j)) <= N / 2) {
        cout << "D";
      } else
        cout << '*';
    }
    cout << '\n';
  }
  return 0;
}

### 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;
template <class T>
inline void smax(T &x, T y) {
  x = max((x), (y));
}
template <class T>
inline void smin(T &x, T y) {
  x = min((x), (y));
}
inline void scanInt(int &x) {
  register char c = getchar();
  x = 0;
  bool neg = false;
  while (c < 48 || c > 57) {
    if (c == '-') neg = true;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x *= -1;
}
int main(int argc, const char *argv[]) {
  int N;
  scanInt(N);
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if ((abs(i - N / 2) + abs(N / 2 - j)) <= N / 2) {
        cout << "D";
      } else
        cout << '*';
    }
    cout << '\n';
  }
  return 0;
}

```