#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i) {
    if (i != v.begin()) os << ", ";
    os << *i;
  }
  os << "]";
  return os;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = 0; j < (int)(n); ++j)
      printf("%c", (abs(n / 2 - i) + abs(n / 2 - j) <= n / 2) ? 'D' : '*');
    printf("\n");
  }
}

### Prompt

Generate 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;
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i) {
    if (i != v.begin()) os << ", ";
    os << *i;
  }
  os << "]";
  return os;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = 0; j < (int)(n); ++j)
      printf("%c", (abs(n / 2 - i) + abs(n / 2 - j) <= n / 2) ? 'D' : '*');
    printf("\n");
  }
}

```