#include <bits/stdc++.h>
using namespace std;
inline long long int inp() {
  long long int n = 0, s = 1;
  char c = getchar();
  while (c < '0') {
    if (c == '-') s = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') n = (n << 3) + (n << 1) + c - '0', c = getchar();
  return n * s;
}
int n, d;
char str[107][107];
int main() {
  n = inp();
  d = (n >> 1);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) str[i][j] = '*';
  for (int i = 0; i <= d; ++i)
    for (int j = (d - i); j <= d; ++j) str[i][j] = 'D';
  for (int i = 0; i <= d; ++i)
    for (int j = d; j <= (d + i); ++j) str[i][j] = 'D';
  for (int i = d + 1; i < n; ++i)
    for (int j = 0; j < n; ++j) str[i][j] = str[n - 1 - i][j];
  for (int i = 0; i < n; ++i) cout << str[i] << endl;
  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;
inline long long int inp() {
  long long int n = 0, s = 1;
  char c = getchar();
  while (c < '0') {
    if (c == '-') s = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') n = (n << 3) + (n << 1) + c - '0', c = getchar();
  return n * s;
}
int n, d;
char str[107][107];
int main() {
  n = inp();
  d = (n >> 1);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) str[i][j] = '*';
  for (int i = 0; i <= d; ++i)
    for (int j = (d - i); j <= d; ++j) str[i][j] = 'D';
  for (int i = 0; i <= d; ++i)
    for (int j = d; j <= (d + i); ++j) str[i][j] = 'D';
  for (int i = d + 1; i < n; ++i)
    for (int j = 0; j < n; ++j) str[i][j] = str[n - 1 - i][j];
  for (int i = 0; i < n; ++i) cout << str[i] << endl;
  return 0;
}

```