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

### 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 x;
  scanf("%d", &x);
  int a;
  a = x / 2;
  int b = a;
  for (int u = 0; u < b; u++) {
    for (int i = 0; i < a; i++) {
      printf("*");
    }
    for (int j = 0; j < (x - a * 2); j++) {
      printf("D");
    }
    for (int i = 0; i < a; i++) {
      printf("*");
    }
    printf("\n");
    a = a - 1;
  }
  for (int h = 0; h < x; h++) {
    printf("D");
  }
  a = a + 1;
  printf("\n");
  for (int u = 0; u < b; u++) {
    for (int i = 0; i < a; i++) {
      printf("*");
    }
    for (int j = 0; j < (x - a * 2); j++) {
      printf("D");
    }
    for (int i = 0; i < a; i++) {
      printf("*");
    }
    printf("\n");
    a = a + 1;
  }
}

```