#include <bits/stdc++.h>
using namespace std;
char mat[105][105];
int n;
int main() {
  cin >> n;
  if (n == 1) {
    cout << "D" << endl;
    return 0;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) mat[i][j] = '*';
  int a = 1;
  int b = n;
  int k = 0;
  n = n / 2 + 1;
  while (n--) {
    for (int i = 0; i < a; i++) mat[k][n + i] = 'D';
    a += 2;
    k++;
  }
  for (int i = 0; i < b; i++) mat[a][i] = 'D';
  for (int i = 0; i < b / 2 + 1; i++) {
    for (int j = 0; j < b; j++) cout << mat[i][j];
    cout << endl;
  }
  for (int i = b / 2 - 1; i >= 0; i--) {
    for (int j = 0; j < b; j++) cout << mat[i][j];
    cout << endl;
  }
}

### 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;
char mat[105][105];
int n;
int main() {
  cin >> n;
  if (n == 1) {
    cout << "D" << endl;
    return 0;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) mat[i][j] = '*';
  int a = 1;
  int b = n;
  int k = 0;
  n = n / 2 + 1;
  while (n--) {
    for (int i = 0; i < a; i++) mat[k][n + i] = 'D';
    a += 2;
    k++;
  }
  for (int i = 0; i < b; i++) mat[a][i] = 'D';
  for (int i = 0; i < b / 2 + 1; i++) {
    for (int j = 0; j < b; j++) cout << mat[i][j];
    cout << endl;
  }
  for (int i = b / 2 - 1; i >= 0; i--) {
    for (int j = 0; j < b; j++) cout << mat[i][j];
    cout << endl;
  }
}

```