#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  string x, r;
  cin >> x;
  bool barra = false;
  for (auto &i : x) {
    if (i == '/') {
      if (barra) continue;
      r += '/';
      barra = true;
    } else {
      r += i;
      barra = false;
    }
  }
  if (r[r.size() - 1] == '/' and r.size() > 1 and r[r.size() - 2] != '/')
    r.erase(r.size() - 1, 1);
  cout << r << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  string x, r;
  cin >> x;
  bool barra = false;
  for (auto &i : x) {
    if (i == '/') {
      if (barra) continue;
      r += '/';
      barra = true;
    } else {
      r += i;
      barra = false;
    }
  }
  if (r[r.size() - 1] == '/' and r.size() > 1 and r[r.size() - 2] != '/')
    r.erase(r.size() - 1, 1);
  cout << r << '\n';
  return 0;
}

```