#include <bits/stdc++.h>
using namespace std;
int n;
char g[103][103];
int vis[103][103];
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
void expand(int x, int y, int k) {
  g[x][y] = 'D';
  vis[x][y] = k;
  for (int i = 0; i < 4; i++) {
    if (k < n / 2 && vis[x + dx[i]][y + dy[i]] > k + 1) {
      expand(x + dx[i], y + dy[i], k + 1);
    }
  }
  return;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      g[i][j] = '*';
      vis[i][j] = 1e9;
    }
  }
  expand(n / 2 + 1, n / 2 + 1, 0);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cout << g[i][j];
    }
    cout << endl;
  }
}

### Prompt

Please provide a cpp coded solution 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 n;
char g[103][103];
int vis[103][103];
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
void expand(int x, int y, int k) {
  g[x][y] = 'D';
  vis[x][y] = k;
  for (int i = 0; i < 4; i++) {
    if (k < n / 2 && vis[x + dx[i]][y + dy[i]] > k + 1) {
      expand(x + dx[i], y + dy[i], k + 1);
    }
  }
  return;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      g[i][j] = '*';
      vis[i][j] = 1e9;
    }
  }
  expand(n / 2 + 1, n / 2 + 1, 0);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cout << g[i][j];
    }
    cout << endl;
  }
}

```