#include <bits/stdc++.h>
using namespace std;
const long long MAX = 1e6 + 1;
long long x, a, cnt, k, y, z, ans, n, m;
vector<long long> v;
pair<long long, long long> p;
set<long long> st;
string s;
char ch[1001][1001];
bool e[MAX], me;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0, j = n / 2, k = 1; i <= (n - 1) / 2; i++) {
    for (int x = 0; x < j; x++) {
      cout << '*';
    }
    for (int x = 0; x < k; x++) {
      cout << 'D';
    }
    for (int x = 0; x < j; x++) {
      cout << '*';
    }
    j--;
    k += 2;
    cout << '\n';
  }
  for (int i = 0, j = 1, k = n - 2; i < (n - 1) / 2; i++) {
    for (int x = 0; x < j; x++) {
      cout << '*';
    }
    for (int x = 0; x < k; x++) {
      cout << 'D';
    }
    for (int x = 0; x < j; x++) {
      cout << '*';
    }
    j++;
    k -= 2;
    cout << '\n';
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long MAX = 1e6 + 1;
long long x, a, cnt, k, y, z, ans, n, m;
vector<long long> v;
pair<long long, long long> p;
set<long long> st;
string s;
char ch[1001][1001];
bool e[MAX], me;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0, j = n / 2, k = 1; i <= (n - 1) / 2; i++) {
    for (int x = 0; x < j; x++) {
      cout << '*';
    }
    for (int x = 0; x < k; x++) {
      cout << 'D';
    }
    for (int x = 0; x < j; x++) {
      cout << '*';
    }
    j--;
    k += 2;
    cout << '\n';
  }
  for (int i = 0, j = 1, k = n - 2; i < (n - 1) / 2; i++) {
    for (int x = 0; x < j; x++) {
      cout << '*';
    }
    for (int x = 0; x < k; x++) {
      cout << 'D';
    }
    for (int x = 0; x < j; x++) {
      cout << '*';
    }
    j++;
    k -= 2;
    cout << '\n';
  }
}

```