#include <bits/stdc++.h>
using namespace std;
int n, a[10000005], c1, c2;
int main() {
  cin >> n;
  a[1] = 1;
  for (int i = 2; i <= n / 2 + 1; ++i) {
    a[i] = a[i - 1] + 2;
  }
  int c = n / 2 + 1;
  for (int i = n / 2 + 2; i <= n; i++) {
    c--;
    a[i] = a[c];
    if (i == n) a[i] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= (n - a[i]) / 2; ++j) {
      cout << '*';
    }
    for (int j = 1; j <= a[i]; ++j) cout << 'D';
    for (int j = 1; j <= (n - a[i]) / 2; ++j) cout << '*';
    cout << endl;
  }
}

### 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;
int n, a[10000005], c1, c2;
int main() {
  cin >> n;
  a[1] = 1;
  for (int i = 2; i <= n / 2 + 1; ++i) {
    a[i] = a[i - 1] + 2;
  }
  int c = n / 2 + 1;
  for (int i = n / 2 + 2; i <= n; i++) {
    c--;
    a[i] = a[c];
    if (i == n) a[i] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= (n - a[i]) / 2; ++j) {
      cout << '*';
    }
    for (int j = 1; j <= a[i]; ++j) cout << 'D';
    for (int j = 1; j <= (n - a[i]) / 2; ++j) cout << '*';
    cout << endl;
  }
}

```