#include <bits/stdc++.h>
using namespace std;
int n, m, i, j, cow = 0, sum = 0, k, l, x, y, ar[10000], ar2[10000];
int main() {
  scanf("%d", &n);
  m = (n - 1) / 2;
  x = 1;
  for (i = 1; i <= n; i += 2) {
    for (k = m; k >= x; k--) {
      printf("*");
    }
    for (j = 1; j <= i; j++) {
      printf("D");
    }
    for (k = m; k >= x; k--) {
      printf("*");
    }
    x = x + 1;
    cout << endl;
  }
  cow = 1;
  for (i = n - 2; i >= 1; i -= 2) {
    for (k = 1; k <= cow; k++) {
      printf("*");
    }
    for (j = 1; j <= i; j++) {
      printf("D");
    }
    for (k = 1; k <= cow; k++) {
      printf("*");
    }
    cow += 1;
    cout << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int n, m, i, j, cow = 0, sum = 0, k, l, x, y, ar[10000], ar2[10000];
int main() {
  scanf("%d", &n);
  m = (n - 1) / 2;
  x = 1;
  for (i = 1; i <= n; i += 2) {
    for (k = m; k >= x; k--) {
      printf("*");
    }
    for (j = 1; j <= i; j++) {
      printf("D");
    }
    for (k = m; k >= x; k--) {
      printf("*");
    }
    x = x + 1;
    cout << endl;
  }
  cow = 1;
  for (i = n - 2; i >= 1; i -= 2) {
    for (k = 1; k <= cow; k++) {
      printf("*");
    }
    for (j = 1; j <= i; j++) {
      printf("D");
    }
    for (k = 1; k <= cow; k++) {
      printf("*");
    }
    cow += 1;
    cout << endl;
  }
  return 0;
}

```