#include <bits/stdc++.h>
char mat[105][105];
void func(int n) {
  int i, j;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) mat[i][j] = '*';
  }
  int mid = (n + 1) / 2;
  int count = 0;
  for (i = 1; i <= mid; i++) {
    mat[i][mid] = 'D';
    for (j = 1; j <= count; j++) {
      mat[i][mid - j] = 'D';
      mat[i][mid + j] = 'D';
    }
    count++;
  }
  for (i = mid + 1; i <= n; i++) {
    for (j = 1; j <= n; j++) mat[i][j] = mat[mid - (i - mid)][j];
  }
}
int main(void) {
  int i, j, n;
  scanf("%d", &n);
  func(n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) printf("%c", mat[i][j]);
    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>
char mat[105][105];
void func(int n) {
  int i, j;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) mat[i][j] = '*';
  }
  int mid = (n + 1) / 2;
  int count = 0;
  for (i = 1; i <= mid; i++) {
    mat[i][mid] = 'D';
    for (j = 1; j <= count; j++) {
      mat[i][mid - j] = 'D';
      mat[i][mid + j] = 'D';
    }
    count++;
  }
  for (i = mid + 1; i <= n; i++) {
    for (j = 1; j <= n; j++) mat[i][j] = mat[mid - (i - mid)][j];
  }
}
int main(void) {
  int i, j, n;
  scanf("%d", &n);
  func(n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) printf("%c", mat[i][j]);
    printf("\n");
  }
  return 0;
}

```