#include <bits/stdc++.h>
using namespace std;
vector<bool> used;
vector<vector<double> > g;
vector<double> qw;
int n, av, k, m;
bool isp = 1;
double const kos = sqrt(2) * 100;
class point {
 public:
  int x, y;
  point(int x, int y) : x(x), y(y) {}
} hajimari(0, 0);
int main() {
  cin >> n;
  for (int i = 1; i <= (n / 2) + 1; i++) {
    int s = ((i - 1) * 2) + 1;
    int fst = (n - s) / 2;
    int j = 1;
    for (j = 1; j <= fst; j++) cout << '*';
    for (j = 1; j <= s; j++) cout << 'D';
    for (j = 1; j <= fst; j++) cout << '*';
    cout << endl;
  }
  for (int i = (n / 2); i >= 1; i--) {
    int s = ((i - 1) * 2) + 1;
    int fst = (n - s) / 2;
    int j = 1;
    for (j = 1; j <= fst; j++) cout << '*';
    for (j = 1; j <= s; j++) cout << 'D';
    for (j = 1; j <= fst; j++) cout << '*';
    cout << endl;
  }
  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;
vector<bool> used;
vector<vector<double> > g;
vector<double> qw;
int n, av, k, m;
bool isp = 1;
double const kos = sqrt(2) * 100;
class point {
 public:
  int x, y;
  point(int x, int y) : x(x), y(y) {}
} hajimari(0, 0);
int main() {
  cin >> n;
  for (int i = 1; i <= (n / 2) + 1; i++) {
    int s = ((i - 1) * 2) + 1;
    int fst = (n - s) / 2;
    int j = 1;
    for (j = 1; j <= fst; j++) cout << '*';
    for (j = 1; j <= s; j++) cout << 'D';
    for (j = 1; j <= fst; j++) cout << '*';
    cout << endl;
  }
  for (int i = (n / 2); i >= 1; i--) {
    int s = ((i - 1) * 2) + 1;
    int fst = (n - s) / 2;
    int j = 1;
    for (j = 1; j <= fst; j++) cout << '*';
    for (j = 1; j <= s; j++) cout << 'D';
    for (j = 1; j <= fst; j++) cout << '*';
    cout << endl;
  }
  return 0;
}

```