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

### 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, x, y, z;
  scanf("%d", &n);
  x = n / 2;
  y = x;
  for (i = 1; i <= n / 2 + 1; i++) {
    while (y != 0) {
      printf("*");
      y--;
    }
    for (k = 1; k <= i * 2 - 1; k++) printf("D");
    y = x;
    while (y != 0) {
      printf("*");
      y--;
    }
    x--;
    y = x;
    printf("\n");
  }
  x = 1;
  y = x;
  for (i = n / 2; i >= 1; i--) {
    while (y != 0) {
      printf("*");
      y--;
    }
    for (j = 1; j <= i * 2 - 1; j++) {
      printf("D");
    }
    y = x;
    while (y != 0) {
      printf("*");
      y--;
    }
    x += 1;
    y = x;
    printf("\n");
  }
  return 0;
}

```