#include <bits/stdc++.h>
using namespace std;
long long i, m, j, n, k, y, ans, a[111];
char x[105][105];
void l(int o, int p, int q) {
  if (o > 0) {
    x[q][p] = 'D';
    --p;
    --o;
    l(o, p, q);
  }
}
void r(int oo, int pp, int qq) {
  if (oo > 0) {
    x[qq][pp] = 'D';
    ++pp;
    --oo;
    r(oo, pp, qq);
  }
}
int main() {
  cin >> n;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) x[i][j] = '*';
  k = n / 2;
  for (i = 1; i <= n; ++i) {
    x[i][k + 1] = 'D';
    l(y, k, i);
    r(y, k + 2, i);
    if (i < (n / 2) + 1) y = y + 1;
    if (i >= (n / 2) + 1) y = y - 1;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j) cout << x[i][j];
    cout << '\n';
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 i, m, j, n, k, y, ans, a[111];
char x[105][105];
void l(int o, int p, int q) {
  if (o > 0) {
    x[q][p] = 'D';
    --p;
    --o;
    l(o, p, q);
  }
}
void r(int oo, int pp, int qq) {
  if (oo > 0) {
    x[qq][pp] = 'D';
    ++pp;
    --oo;
    r(oo, pp, qq);
  }
}
int main() {
  cin >> n;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) x[i][j] = '*';
  k = n / 2;
  for (i = 1; i <= n; ++i) {
    x[i][k + 1] = 'D';
    l(y, k, i);
    r(y, k + 2, i);
    if (i < (n / 2) + 1) y = y + 1;
    if (i >= (n / 2) + 1) y = y - 1;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j) cout << x[i][j];
    cout << '\n';
  }
}

```