#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e6 + 10;
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    int mid = n / 2 + 1;
    int len = 0;
    for (int i = 1; i <= n / 2 + 1; i++) {
      for (int j = 1; j < mid; j++) printf("*");
      for (int j = mid; j <= mid + len; j++) printf("D");
      for (int j = mid + len + 1; j <= n; j++) printf("*");
      printf("\n");
      mid -= 1;
      len += 2;
    }
    mid += 2;
    len -= 4;
    for (int i = n / 2 + 2; i <= n; i++) {
      for (int j = 1; j < mid; j++) printf("*");
      for (int j = mid; j <= mid + len; j++) printf("D");
      for (int j = mid + len + 1; j <= n; j++) printf("*");
      printf("\n");
      mid += 1;
      len -= 2;
    }
  }
  return 0;
}

### 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;
const int MAX = 1e6 + 10;
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    int mid = n / 2 + 1;
    int len = 0;
    for (int i = 1; i <= n / 2 + 1; i++) {
      for (int j = 1; j < mid; j++) printf("*");
      for (int j = mid; j <= mid + len; j++) printf("D");
      for (int j = mid + len + 1; j <= n; j++) printf("*");
      printf("\n");
      mid -= 1;
      len += 2;
    }
    mid += 2;
    len -= 4;
    for (int i = n / 2 + 2; i <= n; i++) {
      for (int j = 1; j < mid; j++) printf("*");
      for (int j = mid; j <= mid + len; j++) printf("D");
      for (int j = mid + len + 1; j <= n; j++) printf("*");
      printf("\n");
      mid += 1;
      len -= 2;
    }
  }
  return 0;
}

```