#include <bits/stdc++.h>
using namespace std;
;
int main() {
  bool d1 = false, d2 = true;
  vector<int> vec1, vec2, vec3;
  long long int n, m, a, total = 0, mid, itr, itr2, mid2;
  cin >> n;
  mid = n / 2;
  mid2 = mid;
  itr = 1;
  itr2 = itr;
  for (int i = (0); i < (n); i++) {
    itr2 = itr;
    for (int j = (0); j < (n); j++) {
      if (j == mid) {
        d1 = true;
      }
      if (itr2 == 0) {
        d1 = false;
      }
      if (d1) {
        cout << "D";
        itr2--;
      } else {
        cout << "*";
      }
    }
    if (i == mid2) {
      d2 = false;
    }
    if (d2) {
      itr = itr + 2;
      mid--;
    } else {
      itr = itr - 2;
      mid++;
    }
    d1 = false;
    cout << endl;
  }
  return 0;
}

### 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;
;
int main() {
  bool d1 = false, d2 = true;
  vector<int> vec1, vec2, vec3;
  long long int n, m, a, total = 0, mid, itr, itr2, mid2;
  cin >> n;
  mid = n / 2;
  mid2 = mid;
  itr = 1;
  itr2 = itr;
  for (int i = (0); i < (n); i++) {
    itr2 = itr;
    for (int j = (0); j < (n); j++) {
      if (j == mid) {
        d1 = true;
      }
      if (itr2 == 0) {
        d1 = false;
      }
      if (d1) {
        cout << "D";
        itr2--;
      } else {
        cout << "*";
      }
    }
    if (i == mid2) {
      d2 = false;
    }
    if (d2) {
      itr = itr + 2;
      mid--;
    } else {
      itr = itr - 2;
      mid++;
    }
    d1 = false;
    cout << endl;
  }
  return 0;
}

```