#include <bits/stdc++.h>
int main() {
  int n, i, j, k, l, m, o, p;
  scanf("%d", &n);
  for (i = 0; i <= n / 2; i++) {
    for (j = 1; j <= n / 2 - i; j++) {
      printf("*");
    }
    for (k = 1; k <= i + 1; k++) {
      printf("D");
    }
    for (m = 0; m < i; m++) {
      printf("D");
    }
    for (l = 0; l < n / 2 - i; l++) {
      printf("*");
    }
    printf("\n");
  }
  for (i = 0; i < n / 2; i++) {
    for (j = 0; j <= i; j++) {
      printf("*");
    }
    for (k = 1; k <= n / 2 - i; k++) {
      printf("D");
    }
    for (m = i; m < n / 2 - 1; m++) {
      printf("D");
    }
    for (l = 0; l <= i; l++) {
      printf("*");
    }
    printf("\n");
  }
}

### Prompt

Create a solution in Cpp for 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, j, k, l, m, o, p;
  scanf("%d", &n);
  for (i = 0; i <= n / 2; i++) {
    for (j = 1; j <= n / 2 - i; j++) {
      printf("*");
    }
    for (k = 1; k <= i + 1; k++) {
      printf("D");
    }
    for (m = 0; m < i; m++) {
      printf("D");
    }
    for (l = 0; l < n / 2 - i; l++) {
      printf("*");
    }
    printf("\n");
  }
  for (i = 0; i < n / 2; i++) {
    for (j = 0; j <= i; j++) {
      printf("*");
    }
    for (k = 1; k <= n / 2 - i; k++) {
      printf("D");
    }
    for (m = i; m < n / 2 - 1; m++) {
      printf("D");
    }
    for (l = 0; l <= i; l++) {
      printf("*");
    }
    printf("\n");
  }
}

```