#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC optimize("fast-math")
using namespace std;
const int INF = 1e9;
const long long INFll = 2e18;
const int BASE1 = 179;
const int BASE2 = 653;
const long long MOD = 1e9 + 7;
const int MAXN = 2e5;
const long double PI = 3.1415926535;
const long double EPS = 1e-10;
void solve() {
  int n;
  cin >> n;
  vector<vector<char>> dop(n, vector<char>(n, '*'));
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (n); j++) {
      if (abs(i - n / 2) + abs(j - n / 2) <= n / 2) dop[i][j] = 'D';
    }
  }
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (n); j++) {
      cout << dop[i][j];
    }
    cout << "\n";
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(15);
  int t = 1;
  while (t--) solve();
  return 0;
}

### 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>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC optimize("fast-math")
using namespace std;
const int INF = 1e9;
const long long INFll = 2e18;
const int BASE1 = 179;
const int BASE2 = 653;
const long long MOD = 1e9 + 7;
const int MAXN = 2e5;
const long double PI = 3.1415926535;
const long double EPS = 1e-10;
void solve() {
  int n;
  cin >> n;
  vector<vector<char>> dop(n, vector<char>(n, '*'));
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (n); j++) {
      if (abs(i - n / 2) + abs(j - n / 2) <= n / 2) dop[i][j] = 'D';
    }
  }
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (n); j++) {
      cout << dop[i][j];
    }
    cout << "\n";
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(15);
  int t = 1;
  while (t--) solve();
  return 0;
}

```