#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, mid;
  while (cin >> n) {
    mid = n / 2;
    char M[n + 1][n + 1];
    int x = mid;
    int y = mid;
    for (int i = 0; i <= mid; ++i) {
      for (int j = 0; j < n; ++j) {
        if (j >= x and j <= y)
          M[i][j] = 'D';
        else
          M[i][j] = '*';
      }
      x--;
      y++;
    }
    x = 0;
    y = n - 1;
    for (int i = mid; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (j >= x and j <= y)
          M[i][j] = 'D';
        else
          M[i][j] = '*';
      }
      x++;
      y--;
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        cout << M[i][j];
      }
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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() {
  int n, mid;
  while (cin >> n) {
    mid = n / 2;
    char M[n + 1][n + 1];
    int x = mid;
    int y = mid;
    for (int i = 0; i <= mid; ++i) {
      for (int j = 0; j < n; ++j) {
        if (j >= x and j <= y)
          M[i][j] = 'D';
        else
          M[i][j] = '*';
      }
      x--;
      y++;
    }
    x = 0;
    y = n - 1;
    for (int i = mid; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (j >= x and j <= y)
          M[i][j] = 'D';
        else
          M[i][j] = '*';
      }
      x++;
      y--;
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        cout << M[i][j];
      }
      cout << endl;
    }
  }
  return 0;
}

```