#include <bits/stdc++.h>
using namespace std;
char s[128];
int main() {
  scanf("%s", s);
  bool rt = true;
  for (char *cpt = s; *cpt;) {
    if (*cpt == '/') {
      while (*cpt == '/') cpt++;
      if (rt || *cpt != '\0') putchar('/');
      rt = false;
    } else
      putchar(*(cpt++));
  }
  putchar('\n');
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
char s[128];
int main() {
  scanf("%s", s);
  bool rt = true;
  for (char *cpt = s; *cpt;) {
    if (*cpt == '/') {
      while (*cpt == '/') cpt++;
      if (rt || *cpt != '\0') putchar('/');
      rt = false;
    } else
      putchar(*(cpt++));
  }
  putchar('\n');
  return 0;
}

```