#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  cin >> s;
  int x = 0;
  int print = 0;
  for (int i = 0; i < s.size(); i++) {
    if (x == 1 && s[i] != '/') {
      cout << '/' << s[i];
      print = 1;
    } else if (s[i] != '/') {
      cout << s[i];
      print = 1;
    }
    if (s[i] == '/')
      x = 1;
    else
      x = 0;
  }
  if (!print) cout << '/';
  cout << endl;
  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;
int main() {
  string s;
  cin >> s;
  int x = 0;
  int print = 0;
  for (int i = 0; i < s.size(); i++) {
    if (x == 1 && s[i] != '/') {
      cout << '/' << s[i];
      print = 1;
    } else if (s[i] != '/') {
      cout << s[i];
      print = 1;
    }
    if (s[i] == '/')
      x = 1;
    else
      x = 0;
  }
  if (!print) cout << '/';
  cout << endl;
  return 0;
}

```