#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9;
const long long MAXN = 1e2 + 69;
const long long MOD = 1e9 + 7;
int n, cnt;
char arr[MAXN][MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  cnt = 1, memset(arr, '*', sizeof arr);
  for (int i = 0; i < n / 2; i++) {
    for (int j = 0; j < cnt; j++) {
      arr[i][n / 2 + j - cnt / 2] = 'D';
    }
    cnt += 2;
  }
  for (int i = 0; i < n / 2; i++) {
    for (int j = 0; j < n; j++) cout << arr[i][j];
    cout << "\n";
  }
  for (int i = 0; i < n; i++) cout << "D";
  cout << "\n";
  for (int i = n / 2 - 1; i >= 0; i--) {
    for (int j = 0; j < n; j++) cout << arr[i][j];
    cout << "\n";
  }
  return cout << '\n', 0;
}

### Prompt

Your challenge is to write 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;
const long long INF = 1e9;
const long long MAXN = 1e2 + 69;
const long long MOD = 1e9 + 7;
int n, cnt;
char arr[MAXN][MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  cnt = 1, memset(arr, '*', sizeof arr);
  for (int i = 0; i < n / 2; i++) {
    for (int j = 0; j < cnt; j++) {
      arr[i][n / 2 + j - cnt / 2] = 'D';
    }
    cnt += 2;
  }
  for (int i = 0; i < n / 2; i++) {
    for (int j = 0; j < n; j++) cout << arr[i][j];
    cout << "\n";
  }
  for (int i = 0; i < n; i++) cout << "D";
  cout << "\n";
  for (int i = n / 2 - 1; i >= 0; i--) {
    for (int j = 0; j < n; j++) cout << arr[i][j];
    cout << "\n";
  }
  return cout << '\n', 0;
}

```