#include <bits/stdc++.h>
using namespace std;
long long int modpow(long long int a, long long int n, long long int temp) {
  long long int res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0)
    return (b);
  else
    return (gcd(b % a, a));
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, i, j;
  cin >> n;
  int star = n / 2, d = 1;
  for (i = 0; i < n / 2; i++) {
    for (j = 0; j < star; j++) cout << "*";
    for (j = 0; j < d; j++) cout << "D";
    for (j = 0; j < star; j++) cout << "*";
    cout << "\n";
    star--;
    d += 2;
  }
  for (i = 0; i < n; i++) cout << "D";
  cout << "\n";
  for (i = 0; i < n / 2; i++) {
    star++;
    d -= 2;
    for (j = 0; j < star; j++) cout << "*";
    for (j = 0; j < d; j++) cout << "D";
    for (j = 0; j < star; j++) cout << "*";
    cout << "\n";
  }
  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;
long long int modpow(long long int a, long long int n, long long int temp) {
  long long int res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0)
    return (b);
  else
    return (gcd(b % a, a));
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, i, j;
  cin >> n;
  int star = n / 2, d = 1;
  for (i = 0; i < n / 2; i++) {
    for (j = 0; j < star; j++) cout << "*";
    for (j = 0; j < d; j++) cout << "D";
    for (j = 0; j < star; j++) cout << "*";
    cout << "\n";
    star--;
    d += 2;
  }
  for (i = 0; i < n; i++) cout << "D";
  cout << "\n";
  for (i = 0; i < n / 2; i++) {
    star++;
    d -= 2;
    for (j = 0; j < star; j++) cout << "*";
    for (j = 0; j < d; j++) cout << "D";
    for (j = 0; j < star; j++) cout << "*";
    cout << "\n";
  }
  return 0;
}

```