#include <bits/stdc++.h>
using namespace std;
const int MAXN1 = 5 * 1e7 + 100;
const int MAXN2 = 5 * 1e3 + 100;
struct xx {
  int st;
  int nd;
};
bool cp(xx a, xx b) { return a.st > b.nd; }
priority_queue<int, vector<int>, greater<int> > qu;
priority_queue<int, vector<int>, less<int> > qd;
int main(void) {
  int n;
  while (cin >> n) {
    for (int i = 0; i < (n + 1) / 2; i++) {
      for (int j = 1; j <= (n - 1) / 2 - i; j++) cout << "*";
      for (int j = 1; j <= i * 2 + 1; j++) cout << "D";
      for (int j = 1; j <= (n - 1) / 2 - i; j++) cout << "*";
      cout << endl;
    }
    for (int i = 1, k = n - 2; i <= (n + 1) / 2 - 1; i++, k -= 2) {
      for (int j = 1; j <= i; j++) cout << "*";
      for (int j = k; j >= 1; j--) cout << "D";
      for (int j = 1; j <= i; j++) cout << "*";
      cout << endl;
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const int MAXN1 = 5 * 1e7 + 100;
const int MAXN2 = 5 * 1e3 + 100;
struct xx {
  int st;
  int nd;
};
bool cp(xx a, xx b) { return a.st > b.nd; }
priority_queue<int, vector<int>, greater<int> > qu;
priority_queue<int, vector<int>, less<int> > qd;
int main(void) {
  int n;
  while (cin >> n) {
    for (int i = 0; i < (n + 1) / 2; i++) {
      for (int j = 1; j <= (n - 1) / 2 - i; j++) cout << "*";
      for (int j = 1; j <= i * 2 + 1; j++) cout << "D";
      for (int j = 1; j <= (n - 1) / 2 - i; j++) cout << "*";
      cout << endl;
    }
    for (int i = 1, k = n - 2; i <= (n + 1) / 2 - 1; i++, k -= 2) {
      for (int j = 1; j <= i; j++) cout << "*";
      for (int j = k; j >= 1; j--) cout << "D";
      for (int j = 1; j <= i; j++) cout << "*";
      cout << endl;
    }
  }
  return 0;
}

```