#include <bits/stdc++.h>
using namespace std;
long long int INF = 1e18 + 5;
long long int power(long long int a, long long int n) {
  long long int res = 1;
  if (n & 1) {
    n--;
    res *= a;
  } else {
    n /= 2;
    a = a * a;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
const long long int N = 1e6 + 5;
const long long int MOD = 1e9 + 7;
void solve() {
  list<string> ans;
  long long int n;
  cin >> n;
  string s(n, 'D');
  ans.push_back(s);
  long long int a = 0, b = n - 1;
  for (long long int i = 1; i <= n / 2; i++) {
    s[a] = '*';
    s[b] = '*';
    ans.push_back(s);
    ans.push_front(s);
    a++;
    b--;
  }
  for (auto it : ans) cout << it << "\n";
  return;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  for (long long int i = 1; i <= T; i++) solve();
}

### 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;
long long int INF = 1e18 + 5;
long long int power(long long int a, long long int n) {
  long long int res = 1;
  if (n & 1) {
    n--;
    res *= a;
  } else {
    n /= 2;
    a = a * a;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
const long long int N = 1e6 + 5;
const long long int MOD = 1e9 + 7;
void solve() {
  list<string> ans;
  long long int n;
  cin >> n;
  string s(n, 'D');
  ans.push_back(s);
  long long int a = 0, b = n - 1;
  for (long long int i = 1; i <= n / 2; i++) {
    s[a] = '*';
    s[b] = '*';
    ans.push_back(s);
    ans.push_front(s);
    a++;
    b--;
  }
  for (auto it : ans) cout << it << "\n";
  return;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  for (long long int i = 1; i <= T; i++) solve();
}

```