#include <bits/stdc++.h>
using namespace std;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int min(long long int a, long long int b) {
  if (a < b) return a;
  return b;
}
long long int mypow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<vector<char>> vec(n);
  int i, j;
  int k = (n + 1) / 2;
  int l = k - 1, r = k - 1;
  vector<char> res;
  for (j = 0; j < k; j++) {
    for (i = 0; i < n; i++) {
      if (i >= l && i <= r)
        res.push_back('D');
      else
        res.push_back('*');
    }
    if (j < k - 1) {
      vec[j] = res;
      vec[n - j - 1] = res;
    } else
      vec[j] = res;
    l--;
    r++;
    res.clear();
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) cout << vec[i][j];
    cout << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int min(long long int a, long long int b) {
  if (a < b) return a;
  return b;
}
long long int mypow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<vector<char>> vec(n);
  int i, j;
  int k = (n + 1) / 2;
  int l = k - 1, r = k - 1;
  vector<char> res;
  for (j = 0; j < k; j++) {
    for (i = 0; i < n; i++) {
      if (i >= l && i <= r)
        res.push_back('D');
      else
        res.push_back('*');
    }
    if (j < k - 1) {
      vec[j] = res;
      vec[n - j - 1] = res;
    } else
      vec[j] = res;
    l--;
    r++;
    res.clear();
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) cout << vec[i][j];
    cout << "\n";
  }
  return 0;
}

```