#include <bits/stdc++.h>
using namespace std;
template <class T>
string toStr(const T &x) {
  stringstream s;
  s << x;
  return s.str();
}
template <class T>
int toInt(const T &x) {
  stringstream s;
  s << x;
  int r;
  s >> r;
  return r;
}
int dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  char mat[n][n];
  memset(mat, '*', sizeof mat);
  int k = 1;
  for (int i = 0; i < n; ++i) {
    if (i <= n / 2) {
      int t = (n - k) / 2;
      for (int j = 0; j < k; ++j) {
        mat[i][t++] = 'D';
      }
      if (i == n / 2)
        k -= 2;
      else
        k += 2;
    } else {
      int t = (n - k) / 2;
      for (int j = 0; j < k; ++j) {
        mat[i][t++] = 'D';
      }
      k -= 2;
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cout << mat[i][j];
    }
    cout << '\n';
  }
  return 0;
}

### 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;
template <class T>
string toStr(const T &x) {
  stringstream s;
  s << x;
  return s.str();
}
template <class T>
int toInt(const T &x) {
  stringstream s;
  s << x;
  int r;
  s >> r;
  return r;
}
int dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  char mat[n][n];
  memset(mat, '*', sizeof mat);
  int k = 1;
  for (int i = 0; i < n; ++i) {
    if (i <= n / 2) {
      int t = (n - k) / 2;
      for (int j = 0; j < k; ++j) {
        mat[i][t++] = 'D';
      }
      if (i == n / 2)
        k -= 2;
      else
        k += 2;
    } else {
      int t = (n - k) / 2;
      for (int j = 0; j < k; ++j) {
        mat[i][t++] = 'D';
      }
      k -= 2;
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cout << mat[i][j];
    }
    cout << '\n';
  }
  return 0;
}

```