#include <bits/stdc++.h>
int main() {
  int n, i, g, h, k, f;
  while (scanf("%d", &n) != EOF) {
    for (g = 1; g <= (n + 1) / 2; g++) {
      h = 0;
      k = 0;
      f = (n - (2 * g - 1)) / 2;
      for (i = 0; i < f; i++) {
        printf("*");
        h++;
      }
      for (k = 0; k < n - 2 * h; k++) {
        printf("D");
      }
      for (i = 0; i < h; i++) {
        printf("*");
      }
      printf("\n");
    }
    for (g = 1; g <= n / 2; g++) {
      h = 0;
      for (i = 0; i < g; i++) {
        printf("*");
        h++;
      }
      for (i = 0; i < n - 2 * h; i++) {
        printf("D");
      }
      for (i = 0; i < h; i++) {
        printf("*");
      }
      printf("\n");
    }
  }
  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>
int main() {
  int n, i, g, h, k, f;
  while (scanf("%d", &n) != EOF) {
    for (g = 1; g <= (n + 1) / 2; g++) {
      h = 0;
      k = 0;
      f = (n - (2 * g - 1)) / 2;
      for (i = 0; i < f; i++) {
        printf("*");
        h++;
      }
      for (k = 0; k < n - 2 * h; k++) {
        printf("D");
      }
      for (i = 0; i < h; i++) {
        printf("*");
      }
      printf("\n");
    }
    for (g = 1; g <= n / 2; g++) {
      h = 0;
      for (i = 0; i < g; i++) {
        printf("*");
        h++;
      }
      for (i = 0; i < n - 2 * h; i++) {
        printf("D");
      }
      for (i = 0; i < h; i++) {
        printf("*");
      }
      printf("\n");
    }
  }
  return 0;
}

```