#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e15 + 7;
const long long N = 3e5 + 5;
long long A[N];
void solve() {
  long long i, j, a, b, n, k, first = 0, second = 0;
  cin >> n;
  for (i = 0; i < n / 2 + 1; i++) {
    for (j = 0; j < n / 2 - i; j++) cout << '*';
    for (j = n / 2 - i; j < n / 2 + i + 1; j++) cout << 'D';
    for (j = n / 2 + i + 1; j < n; j++) cout << '*';
    cout << "\n";
  }
  for (i = n / 2 + 1; i < n; i++) {
    for (j = 0; j < i - n / 2; j++) cout << '*';
    for (j = 0; j < n - 2 * (i - n / 2); j++) cout << 'D';
    for (j = 0; j < i - n / 2; j++) cout << '*';
    cout << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, t = 1, j, first;
  while (t--) solve();
}

### 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 long long inf = 1e15 + 7;
const long long N = 3e5 + 5;
long long A[N];
void solve() {
  long long i, j, a, b, n, k, first = 0, second = 0;
  cin >> n;
  for (i = 0; i < n / 2 + 1; i++) {
    for (j = 0; j < n / 2 - i; j++) cout << '*';
    for (j = n / 2 - i; j < n / 2 + i + 1; j++) cout << 'D';
    for (j = n / 2 + i + 1; j < n; j++) cout << '*';
    cout << "\n";
  }
  for (i = n / 2 + 1; i < n; i++) {
    for (j = 0; j < i - n / 2; j++) cout << '*';
    for (j = 0; j < n - 2 * (i - n / 2); j++) cout << 'D';
    for (j = 0; j < i - n / 2; j++) cout << '*';
    cout << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, t = 1, j, first;
  while (t--) solve();
}

```