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

### Prompt

Your task is to create 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, j, k, l, m, x, y;
  scanf("%d", &n);
  x = n / 2, y = 1, m = 1;
  for (i = 1; i < n / 2 + 1; i++) {
    for (j = x; j > 0; j--) {
      printf("*");
    }
    {
      for (k = 1; k <= m; k++) {
        printf("D");
      }
    }
    for (l = x; l > 0; l--) {
      printf("*");
    }
    printf("\n");
    x--;
    y += 2, m += 2;
  }
  for (i = 1; i <= n; i++) {
    printf("D");
  }
  printf("\n");
  m = n - 2;
  for (i = 0; i < n / 2; i++) {
    for (j = 0; j <= i; j++) {
      printf("*");
    }
    for (k = m; k >= 1; k--) {
      printf("D");
    }
    for (l = 0; l <= i; l++) {
      printf("*");
    }
    printf("\n");
    m -= 2;
  }
  return 0;
}

```