#include <bits/stdc++.h>
using namespace std;
void _fill_int(int* p, int val, int rep) {
  int i;
  for (i = 0; i < rep; i++) p[i] = val;
}
void solve() {
  int f, i, j, k, l, x, y;
  cin >> l;
  for (y = 0; y < l; y++) {
    for (x = 0; x < l; x++) {
      if (x < l / 2 - min(y, l - 1 - y) || x > l / 2 + min(y, l - 1 - y))
        (void)printf("*");
      else
        (void)printf("D");
    }
    (void)printf("\n");
  }
}
int main(int argc, char** argv) {
  string s;
  if (argc == 1) ios::sync_with_stdio(false);
  for (int i = 1; i < argc; i++) s += argv[i], s += '\n';
  for (int i = s.size() - 1; i >= 0; i--) ungetc(s[i], stdin);
  solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
void _fill_int(int* p, int val, int rep) {
  int i;
  for (i = 0; i < rep; i++) p[i] = val;
}
void solve() {
  int f, i, j, k, l, x, y;
  cin >> l;
  for (y = 0; y < l; y++) {
    for (x = 0; x < l; x++) {
      if (x < l / 2 - min(y, l - 1 - y) || x > l / 2 + min(y, l - 1 - y))
        (void)printf("*");
      else
        (void)printf("D");
    }
    (void)printf("\n");
  }
}
int main(int argc, char** argv) {
  string s;
  if (argc == 1) ios::sync_with_stdio(false);
  for (int i = 1; i < argc; i++) s += argv[i], s += '\n';
  for (int i = s.size() - 1; i >= 0; i--) ungetc(s[i], stdin);
  solve();
  return 0;
}

```