#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  char A[n + 5][n + 5];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      A[i][j] = 'D';
    }
  }
  int mx = n / 2;
  int lim1 = mx;
  int lim2 = n - mx;
  for (int r = 1; r <= mx; r++) {
    for (int col1 = 1, col2 = n; col1 <= lim1 && col2 >= lim2; col1++, col2--) {
      A[r][col1] = '*';
      A[r][col2] = '*';
    }
    lim1--;
    lim2++;
  }
  lim1 = mx;
  lim2 = n - mx;
  for (int r = n; r >= n - mx; r--) {
    for (int col1 = 1, col2 = n; col1 <= lim1 and col2 >= lim2;
         col1++, col2--) {
      A[r][col1] = '*';
      A[r][col2] = '*';
    }
    lim1--;
    lim2++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cout << A[i][j];
    }
    cout << "\n";
  }
  return 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;
int main() {
  int n;
  cin >> n;
  char A[n + 5][n + 5];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      A[i][j] = 'D';
    }
  }
  int mx = n / 2;
  int lim1 = mx;
  int lim2 = n - mx;
  for (int r = 1; r <= mx; r++) {
    for (int col1 = 1, col2 = n; col1 <= lim1 && col2 >= lim2; col1++, col2--) {
      A[r][col1] = '*';
      A[r][col2] = '*';
    }
    lim1--;
    lim2++;
  }
  lim1 = mx;
  lim2 = n - mx;
  for (int r = n; r >= n - mx; r--) {
    for (int col1 = 1, col2 = n; col1 <= lim1 and col2 >= lim2;
         col1++, col2--) {
      A[r][col1] = '*';
      A[r][col2] = '*';
    }
    lim1--;
    lim2++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cout << A[i][j];
    }
    cout << "\n";
  }
  return 0;
}

```