#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int n, pos;
char a[N][N];
bool diff(int i, int j) { return abs(i - n / 2) + abs(j - n / 2) <= n / 2; }
int main() {
  scanf("%d", &n);
  pos = n / 2;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (diff(i, j))
        a[i][j] = 'D';
      else
        a[i][j] = '*';
    }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) printf("%c", a[i][j]);
    printf("\n");
  }
}

### Prompt

In CPP, your task is to solve 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 N = 110;
int n, pos;
char a[N][N];
bool diff(int i, int j) { return abs(i - n / 2) + abs(j - n / 2) <= n / 2; }
int main() {
  scanf("%d", &n);
  pos = n / 2;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (diff(i, j))
        a[i][j] = 'D';
      else
        a[i][j] = '*';
    }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) printf("%c", a[i][j]);
    printf("\n");
  }
}

```