#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
char s[110][110];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n / 2 + 1; i++) {
      int num = 2 * i - 1;
      int t1 = (n - num) / 2;
      for (int j = 1; j <= t1; j++) s[i][j] = '*';
      for (int j = t1 + 1; j <= t1 + num; j++) s[i][j] = 'D';
      for (int j = t1 + num + 1; j <= n; j++) s[i][j] = '*';
      s[i][n + 1] = '\0';
      if (i != n / 2 + 1) {
        for (int j = 1; j <= n; j++) s[n + 1 - i][j] = s[i][j];
      }
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) printf("%c", s[i][j]);
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const int maxn = 100010;
char s[110][110];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n / 2 + 1; i++) {
      int num = 2 * i - 1;
      int t1 = (n - num) / 2;
      for (int j = 1; j <= t1; j++) s[i][j] = '*';
      for (int j = t1 + 1; j <= t1 + num; j++) s[i][j] = 'D';
      for (int j = t1 + num + 1; j <= n; j++) s[i][j] = '*';
      s[i][n + 1] = '\0';
      if (i != n / 2 + 1) {
        for (int j = 1; j <= n; j++) s[n + 1 - i][j] = s[i][j];
      }
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) printf("%c", s[i][j]);
      printf("\n");
    }
  }
  return 0;
}

```