#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e6 + 123;
const int MOD = (int)1e9 + 7;
const int inf = (int)5e8;
const long long infll = (long long)1e17;
void solve() {
  int n;
  scanf("%d", &n);
  vector<string> ans;
  for (int i = 0; i < n / 2; ++i) {
    string cur = "";
    for (int j = 0; j < i + 1; ++j) {
      cur += "*";
    }
    for (int j = 0; j < n - 2 * (i + 1); ++j) {
      cur += "D";
    }
    for (int j = 0; j < i + 1; ++j) {
      cur += "*";
    }
    ans.push_back(cur);
  }
  deque<string> cur;
  cur.push_back(string(n, 'D'));
  for (int i = 0; i < ans.size(); ++i) {
    cur.push_back(ans[i]);
  }
  for (int i = 0; i < ans.size(); ++i) {
    cur.push_front(ans[i]);
  }
  for (auto i : cur) {
    cout << i << endl;
  }
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
const int N = (int)1e6 + 123;
const int MOD = (int)1e9 + 7;
const int inf = (int)5e8;
const long long infll = (long long)1e17;
void solve() {
  int n;
  scanf("%d", &n);
  vector<string> ans;
  for (int i = 0; i < n / 2; ++i) {
    string cur = "";
    for (int j = 0; j < i + 1; ++j) {
      cur += "*";
    }
    for (int j = 0; j < n - 2 * (i + 1); ++j) {
      cur += "D";
    }
    for (int j = 0; j < i + 1; ++j) {
      cur += "*";
    }
    ans.push_back(cur);
  }
  deque<string> cur;
  cur.push_back(string(n, 'D'));
  for (int i = 0; i < ans.size(); ++i) {
    cur.push_back(ans[i]);
  }
  for (int i = 0; i < ans.size(); ++i) {
    cur.push_front(ans[i]);
  }
  for (auto i : cur) {
    cout << i << endl;
  }
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```