#include <bits/stdc++.h>
using namespace std;
int main() {
  int ok = 1;
  string second, ans;
  cin >> second;
  for (int i = 0; i < (int)(second.size()); i++) {
    if (second[i] != '/')
      ans += second[i], ok = 1;
    else if (second[i] == '/' && ok == 1)
      ok = 0, ans += '/';
  }
  if (ans.back() == '/' && ans.size() > 1) ans.pop_back();
  cout << ans;
}

### Prompt

In Cpp, your task is to solve the following problem:

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() {
  int ok = 1;
  string second, ans;
  cin >> second;
  for (int i = 0; i < (int)(second.size()); i++) {
    if (second[i] != '/')
      ans += second[i], ok = 1;
    else if (second[i] == '/' && ok == 1)
      ok = 0, ans += '/';
  }
  if (ans.back() == '/' && ans.size() > 1) ans.pop_back();
  cout << ans;
}

```