#include <bits/stdc++.h>
char s[110][110];
int main() {
  int i, j, k, l, m, n, p, q, a;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) s[i][j] = '*';
  }
  for (i = 0, a = 1; i < n / 2; i++, a++) {
    k = (2 * a) - 1;
    k -= 1;
    k /= 2;
    p = n / 2;
    for (j = p - 1, l = p + 1, q = 0; q < k; q++, j--, l++) {
      s[i][j] = s[i][l] = 'D';
    }
  }
  for (i = n - 1, a = 1; i > n / 2; i--, a++) {
    k = (2 * a) - 1;
    k -= 1;
    k /= 2;
    p = n / 2;
    for (j = p - 1, l = p + 1, q = 0; q < k; q++, l++, j--) {
      s[i][j] = s[i][l] = 'D';
    }
  }
  for (i = 0; i < n; i++) s[n / 2][i] = s[i][n / 2] = 'D';
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) printf("%c", s[i][j]);
    printf("\n");
  }
  return (0);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 s[110][110];
int main() {
  int i, j, k, l, m, n, p, q, a;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) s[i][j] = '*';
  }
  for (i = 0, a = 1; i < n / 2; i++, a++) {
    k = (2 * a) - 1;
    k -= 1;
    k /= 2;
    p = n / 2;
    for (j = p - 1, l = p + 1, q = 0; q < k; q++, j--, l++) {
      s[i][j] = s[i][l] = 'D';
    }
  }
  for (i = n - 1, a = 1; i > n / 2; i--, a++) {
    k = (2 * a) - 1;
    k -= 1;
    k /= 2;
    p = n / 2;
    for (j = p - 1, l = p + 1, q = 0; q < k; q++, l++, j--) {
      s[i][j] = s[i][l] = 'D';
    }
  }
  for (i = 0; i < n; i++) s[n / 2][i] = s[i][n / 2] = 'D';
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) printf("%c", s[i][j]);
    printf("\n");
  }
  return (0);
}

```