#include <bits/stdc++.h>
using namespace std;
int main() {
  string t;
  cin >> t;
  int c = 0, n = 0;
  if (t == "/") {
    cout << t;
    return 0;
  }
  while (1) {
    for (int i = 0; i <= t.length() - 1; i++)
      if (t[i] == '/' && t[i + 1] == '/') {
        t[i] = ':';
        n++;
      }
    if (n == c)
      break;
    else
      c = n;
  }
  string print = "";
  for (int i = 0; i < t.size(); i++)
    if (t[i] != ':') {
      print.push_back(t[i]);
    };
  if (print == "") print = "/";
  if (print.length() > 1 && print.back() == '/') print.pop_back();
  cout << print << endl;
}

### 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>
using namespace std;
int main() {
  string t;
  cin >> t;
  int c = 0, n = 0;
  if (t == "/") {
    cout << t;
    return 0;
  }
  while (1) {
    for (int i = 0; i <= t.length() - 1; i++)
      if (t[i] == '/' && t[i + 1] == '/') {
        t[i] = ':';
        n++;
      }
    if (n == c)
      break;
    else
      c = n;
  }
  string print = "";
  for (int i = 0; i < t.size(); i++)
    if (t[i] != ':') {
      print.push_back(t[i]);
    };
  if (print == "") print = "/";
  if (print.length() > 1 && print.back() == '/') print.pop_back();
  cout << print << endl;
}

```