#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n;
  cin >> n;
  char a[123][123];
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      a[i][j] = '*';
    }
  }
  long long mid = (n / 2 + 1);
  for (long long i = 1; i <= n / 2 + 1; i++) {
    for (long long j = mid - (i - 1); j <= mid; j++) {
      a[j][i] = 'D';
    }
    for (long long j = mid; j <= mid + (i - 1); j++) {
      a[j][i] = 'D';
    }
  }
  for (long long i = n; i >= n / 2 + 1; i--) {
    for (long long j = mid - (n - i); j <= mid; j++) {
      a[j][i] = 'D';
    }
    for (long long j = mid; j <= mid + (n - i); j++) {
      a[j][i] = 'D';
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      cout << a[i][j];
    }
    cout << "\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;
int main() {
  long long n;
  cin >> n;
  char a[123][123];
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      a[i][j] = '*';
    }
  }
  long long mid = (n / 2 + 1);
  for (long long i = 1; i <= n / 2 + 1; i++) {
    for (long long j = mid - (i - 1); j <= mid; j++) {
      a[j][i] = 'D';
    }
    for (long long j = mid; j <= mid + (i - 1); j++) {
      a[j][i] = 'D';
    }
  }
  for (long long i = n; i >= n / 2 + 1; i--) {
    for (long long j = mid - (n - i); j <= mid; j++) {
      a[j][i] = 'D';
    }
    for (long long j = mid; j <= mid + (n - i); j++) {
      a[j][i] = 'D';
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      cout << a[i][j];
    }
    cout << "\n";
  }
  return 0;
}

```