#include <bits/stdc++.h>
using namespace std;
int ans[104];
int main() {
  int n, i, j;
  scanf("%d", &n);
  i = 1;
  int k = 0;
  while (i <= n) {
    ans[k++] = i;
    i += 2;
  }
  i = n - 2;
  while (i >= 1) {
    ans[k++] = i;
    i -= 2;
  }
  for (i = 0; i < k; i++) {
    int need = n - ans[i];
    int le = need / 2;
    for (j = 0; j < le; j++) printf("*");
    for (j = 0; j < ans[i]; j++) printf("D");
    for (j = 0; j < le; j++) printf("*");
    puts("");
  }
  puts("");
  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>
using namespace std;
int ans[104];
int main() {
  int n, i, j;
  scanf("%d", &n);
  i = 1;
  int k = 0;
  while (i <= n) {
    ans[k++] = i;
    i += 2;
  }
  i = n - 2;
  while (i >= 1) {
    ans[k++] = i;
    i -= 2;
  }
  for (i = 0; i < k; i++) {
    int need = n - ans[i];
    int le = need / 2;
    for (j = 0; j < le; j++) printf("*");
    for (j = 0; j < ans[i]; j++) printf("D");
    for (j = 0; j < le; j++) printf("*");
    puts("");
  }
  puts("");
  return 0;
}

```