#include <bits/stdc++.h>
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    char a[n][n];
    int i = 0, j;
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) a[i][j] = '*';
    }
    int index = n / 2;
    a[0][index] = 'D';
    index--;
    int num = 3;
    for (i = 1; i < n / 2 + 1; i++) {
      int counter = 0;
      if (i != n / 2) {
        for (j = index; counter < num; j++) {
          a[i][j] = 'D';
          counter++;
        }
      } else {
        int k;
        for (k = 0; k < n; k++) a[i][k] = 'D';
      }
      index--;
      num += 2;
    }
    index += 2;
    int num2 = n - 2;
    for (i = n / 2 + 1; i < n - 1; i++) {
      int counter = 0;
      for (j = index; counter < num2; j++) {
        a[i][j] = 'D';
        counter++;
      }
      index++;
      num2 -= 2;
    }
    a[n - 1][n / 2] = 'D';
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (a[i][j] != 'D')
          printf("*");
        else
          printf("%c", a[i][j]);
      }
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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>
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    char a[n][n];
    int i = 0, j;
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) a[i][j] = '*';
    }
    int index = n / 2;
    a[0][index] = 'D';
    index--;
    int num = 3;
    for (i = 1; i < n / 2 + 1; i++) {
      int counter = 0;
      if (i != n / 2) {
        for (j = index; counter < num; j++) {
          a[i][j] = 'D';
          counter++;
        }
      } else {
        int k;
        for (k = 0; k < n; k++) a[i][k] = 'D';
      }
      index--;
      num += 2;
    }
    index += 2;
    int num2 = n - 2;
    for (i = n / 2 + 1; i < n - 1; i++) {
      int counter = 0;
      for (j = index; counter < num2; j++) {
        a[i][j] = 'D';
        counter++;
      }
      index++;
      num2 -= 2;
    }
    a[n - 1][n / 2] = 'D';
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (a[i][j] != 'D')
          printf("*");
        else
          printf("%c", a[i][j]);
      }
      printf("\n");
    }
  }
  return 0;
}

```