#include <bits/stdc++.h>
int main() {
  char s[110];
  int i, j, k = 0, z, lengh, t;
  scanf("%s", s);
  lengh = strlen(s);
  for (i = 0; i < lengh; i++) {
    if (s[i] == '/') {
      for (j = i; j < lengh; j++) {
        if (s[j] != '/') {
          k++;
          break;
        }
      }
      z = j - i - 1;
      t = j;
      for (; j < lengh; j++) s[j - z] = s[j];
      lengh -= z;
    }
  }
  for (i = 0; i < lengh - 1; i++) printf("%c", s[i]);
  if (s[lengh - 1] != '/') {
    printf("%c", s[lengh - 1]);
  }
  if (k == 0) printf("/");
  printf("\n");
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.

A path called normalized if it contains the smallest possible number of characters '/'.

Your task is to transform a given path to the normalized form.

Input

The first line of the input contains only lowercase Latin letters and character '/' — the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty.

Output

The path in normalized form.

Examples

Input

//usr///local//nginx/sbin


Output

/usr/local/nginx/sbin

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  char s[110];
  int i, j, k = 0, z, lengh, t;
  scanf("%s", s);
  lengh = strlen(s);
  for (i = 0; i < lengh; i++) {
    if (s[i] == '/') {
      for (j = i; j < lengh; j++) {
        if (s[j] != '/') {
          k++;
          break;
        }
      }
      z = j - i - 1;
      t = j;
      for (; j < lengh; j++) s[j - z] = s[j];
      lengh -= z;
    }
  }
  for (i = 0; i < lengh - 1; i++) printf("%c", s[i]);
  if (s[lengh - 1] != '/') {
    printf("%c", s[lengh - 1]);
  }
  if (k == 0) printf("/");
  printf("\n");
  return 0;
}

```