#include <bits/stdc++.h>
using namespace std;
long long mini, minj, maxi, maxj, a, b, c, d, e, f, g, h, i, j, k, l, m, n, o,
    p, q, r, s, t, u, v, w, x, y, z, sum, aa[1007], bb[1007];
char c1, c2, c3, c4, c5, c6, c7, c8, c9, c0[507][507];
string s1, s2, s3, s4, s5, s6, s7, s8, s9, s0;
long double d1, d2, d3, d4, d5, d6, d7, d8, d9, d0;
set<char> sc1, sc2;
set<long long> sl1, sl2;
map<long long, long long> mll1, mll2, mll3;
pair<int, int> pr[107];
int main() {
  cin >> n;
  a = -1 * (n / 2);
  for (i = a; i <= n / 2; i++) {
    for (j = a; j <= n / 2; j++) {
      if (abs(i) + abs(j) > n / 2)
        cout << "*";
      else
        cout << "D";
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 mini, minj, maxi, maxj, a, b, c, d, e, f, g, h, i, j, k, l, m, n, o,
    p, q, r, s, t, u, v, w, x, y, z, sum, aa[1007], bb[1007];
char c1, c2, c3, c4, c5, c6, c7, c8, c9, c0[507][507];
string s1, s2, s3, s4, s5, s6, s7, s8, s9, s0;
long double d1, d2, d3, d4, d5, d6, d7, d8, d9, d0;
set<char> sc1, sc2;
set<long long> sl1, sl2;
map<long long, long long> mll1, mll2, mll3;
pair<int, int> pr[107];
int main() {
  cin >> n;
  a = -1 * (n / 2);
  for (i = a; i <= n / 2; i++) {
    for (j = a; j <= n / 2; j++) {
      if (abs(i) + abs(j) > n / 2)
        cout << "*";
      else
        cout << "D";
    }
    cout << endl;
  }
  return 0;
}

```