#include <bits/stdc++.h>
using namespace std;
char str[100 + 5][100 + 5];
int main() {
  int n;
  while (cin >> n) {
    int i, j, k;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        str[i][j] = '*';
      }
    }
    for (i = 1; i <= n / 2 + 1; i++) {
      for (j = n / 2 + 2 - i; j <= n / 2 + i; j++) {
        str[i][j] = 'D';
      }
    }
    for (i = n / 2 + 2; i <= n; i++) {
      for (j = i - n / 2; j <= n - (i - n / 2) + 1; j++) {
        str[i][j] = 'D';
      }
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        printf("%c", str[i][j]);
      }
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
char str[100 + 5][100 + 5];
int main() {
  int n;
  while (cin >> n) {
    int i, j, k;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        str[i][j] = '*';
      }
    }
    for (i = 1; i <= n / 2 + 1; i++) {
      for (j = n / 2 + 2 - i; j <= n / 2 + i; j++) {
        str[i][j] = 'D';
      }
    }
    for (i = n / 2 + 2; i <= n; i++) {
      for (j = i - n / 2; j <= n - (i - n / 2) + 1; j++) {
        str[i][j] = 'D';
      }
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        printf("%c", str[i][j]);
      }
      printf("\n");
    }
  }
  return 0;
}

```