#include <bits/stdc++.h>
using namespace std;
struct edge {
  int from, to, cost;
};
int main() {
  int n;
  string mat[105];
  cin >> n;
  string s;
  for (int i = (0); i < (n); i++) s += '*';
  for (int i = (0); i < (n); i++) mat[i] = s;
  mat[0][n / 2] = 'D';
  mat[n - 1][n / 2] = 'D';
  for (int i = (1); i < (n / 2 + 1); i++) {
    for (int j = (0); j < (n); j++) {
      if (mat[i - 1][j] == 'D') {
        mat[i][j - 1] = mat[i][j] = mat[i][j + 1] = 'D';
      }
    }
  }
  for (int i = n - 2; i > n / 2; i--) {
    for (int j = (0); j < (n); j++) {
      if (mat[i + 1][j] == 'D') {
        mat[i][j - 1] = mat[i][j] = mat[i][j + 1] = 'D';
      }
    }
  }
  for (int i = (0); i < (n); i++) {
    cout << mat[i] << 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;
struct edge {
  int from, to, cost;
};
int main() {
  int n;
  string mat[105];
  cin >> n;
  string s;
  for (int i = (0); i < (n); i++) s += '*';
  for (int i = (0); i < (n); i++) mat[i] = s;
  mat[0][n / 2] = 'D';
  mat[n - 1][n / 2] = 'D';
  for (int i = (1); i < (n / 2 + 1); i++) {
    for (int j = (0); j < (n); j++) {
      if (mat[i - 1][j] == 'D') {
        mat[i][j - 1] = mat[i][j] = mat[i][j + 1] = 'D';
      }
    }
  }
  for (int i = n - 2; i > n / 2; i--) {
    for (int j = (0); j < (n); j++) {
      if (mat[i + 1][j] == 'D') {
        mat[i][j - 1] = mat[i][j] = mat[i][j + 1] = 'D';
      }
    }
  }
  for (int i = (0); i < (n); i++) {
    cout << mat[i] << endl;
  }
  return 0;
}

```