#include <bits/stdc++.h>
const double pi = 3.1415926;
using namespace std;
long long fact(long long n) {
  long long f = 1;
  for (long long i = n; i >= 1; i--) f *= i;
  return f;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int divisor(int n) {
  int sum = 0;
  if (n / sqrt(n) == sqrt(n)) sum++;
  for (int i = 1; i < sqrt(n); i++)
    if (n % i == 0) sum += 2;
  return sum;
}
bool prime(long long n) {
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
bool palindrome_str(string n) {
  int x = 0;
  int m = n.size() / 2 - 1;
  int r = n.size();
  for (int i = 0; i < m; i++)
    if (n[i] == n[r - i - 1]) x++;
  if (x == m) return true;
  return false;
}
bool palindrome_num(int n) {
  int x = 0, a[10000], r = 0;
  while (n) {
    a[r++] = n % 10;
    n /= 10;
  }
  for (int i = 0; i < r / 2; i++)
    if (a[i] == a[r - i - 1]) x++;
  if (x == r / 2) return true;
  return false;
}
void small(string n) {
  for (int i = 0; i < n.size(); i++)
    if (n[i] >= 'A' && n[i] <= 'Z') n[i] += 32;
}
void capital(string n) {
  for (int i = 0; i < n.size(); i++)
    if (n[i] >= 'a' && n[i] <= 'z') n[i] -= 32;
}
long long fast_power(long long n, long long m) {
  if (m == 0) return 0;
  if (m == 1) return n;
  long long x = fast_power(n, m / 2) % 1000000007;
  x = (x * x) % 1000000007;
  if (m % 2 != 0) x *= n;
  x %= 1000000007;
  return x;
}
void fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  fast();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
      if (i + j >= n / 2 && j - i <= n / 2 && i - j <= n / 2 &&
          i + j <= n / 2 * 3)
        cout << "D";
      else
        cout << "*";
    cout << "\n";
  }
}

### Prompt

Your challenge is to write 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>
const double pi = 3.1415926;
using namespace std;
long long fact(long long n) {
  long long f = 1;
  for (long long i = n; i >= 1; i--) f *= i;
  return f;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int divisor(int n) {
  int sum = 0;
  if (n / sqrt(n) == sqrt(n)) sum++;
  for (int i = 1; i < sqrt(n); i++)
    if (n % i == 0) sum += 2;
  return sum;
}
bool prime(long long n) {
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
bool palindrome_str(string n) {
  int x = 0;
  int m = n.size() / 2 - 1;
  int r = n.size();
  for (int i = 0; i < m; i++)
    if (n[i] == n[r - i - 1]) x++;
  if (x == m) return true;
  return false;
}
bool palindrome_num(int n) {
  int x = 0, a[10000], r = 0;
  while (n) {
    a[r++] = n % 10;
    n /= 10;
  }
  for (int i = 0; i < r / 2; i++)
    if (a[i] == a[r - i - 1]) x++;
  if (x == r / 2) return true;
  return false;
}
void small(string n) {
  for (int i = 0; i < n.size(); i++)
    if (n[i] >= 'A' && n[i] <= 'Z') n[i] += 32;
}
void capital(string n) {
  for (int i = 0; i < n.size(); i++)
    if (n[i] >= 'a' && n[i] <= 'z') n[i] -= 32;
}
long long fast_power(long long n, long long m) {
  if (m == 0) return 0;
  if (m == 1) return n;
  long long x = fast_power(n, m / 2) % 1000000007;
  x = (x * x) % 1000000007;
  if (m % 2 != 0) x *= n;
  x %= 1000000007;
  return x;
}
void fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  fast();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
      if (i + j >= n / 2 && j - i <= n / 2 && i - j <= n / 2 &&
          i + j <= n / 2 * 3)
        cout << "D";
      else
        cout << "*";
    cout << "\n";
  }
}

```