#include <bits/stdc++.h>
int a[101][101] = {0};
int main() {
  int i;
  int j;
  int n;
  scanf("%d", &n);
  int count = 0;
  int mid = n / 2;
  for (i = 0; i <= n / 2; i++) {
    int temp = 1;
    a[i][mid] = 1;
    while (temp <= count) {
      a[i][mid + temp] = 1;
      temp++;
    }
    temp = 1;
    a[i][mid] = 1;
    while (temp <= count) {
      a[i][mid - temp] = 1;
      temp++;
    }
    count += 1;
  }
  count = count - 2;
  for (i = n / 2 + 1; i < n; i++) {
    int temp = 1;
    a[i][mid] = 1;
    while (temp <= count) {
      a[i][mid + temp] = 1;
      temp++;
    }
    temp = 1;
    a[i][mid] = 1;
    while (temp <= count) {
      a[i][mid - temp] = 1;
      temp++;
    }
    count -= 1;
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (a[i][j] == 0) {
        printf("*");
      } else
        printf("D");
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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>
int a[101][101] = {0};
int main() {
  int i;
  int j;
  int n;
  scanf("%d", &n);
  int count = 0;
  int mid = n / 2;
  for (i = 0; i <= n / 2; i++) {
    int temp = 1;
    a[i][mid] = 1;
    while (temp <= count) {
      a[i][mid + temp] = 1;
      temp++;
    }
    temp = 1;
    a[i][mid] = 1;
    while (temp <= count) {
      a[i][mid - temp] = 1;
      temp++;
    }
    count += 1;
  }
  count = count - 2;
  for (i = n / 2 + 1; i < n; i++) {
    int temp = 1;
    a[i][mid] = 1;
    while (temp <= count) {
      a[i][mid + temp] = 1;
      temp++;
    }
    temp = 1;
    a[i][mid] = 1;
    while (temp <= count) {
      a[i][mid - temp] = 1;
      temp++;
    }
    count -= 1;
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (a[i][j] == 0) {
        printf("*");
      } else
        printf("D");
    }
    printf("\n");
  }
  return 0;
}

```