#include <bits/stdc++.h>
using namespace std;
static const double EPS = 1e-8;
int main() {
  istream &fin = cin;
  ostream &fout = cout;
  FILE *fpin = stdin;
  FILE *fpout = stdout;
  string s;
  fin >> s;
  bool p = true;
  string res;
  for (int i = s.size() - 1; i >= 0; i--) {
    if (p && s[i] == '/') continue;
    if (s[i] != '/')
      p = false;
    else
      p = true;
    res.push_back(s[i]);
  }
  reverse(res.begin(), res.end());
  if (res.empty()) res = "/";
  fout << res << endl;
}

### Prompt

Please formulate a cpp solution to 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;
static const double EPS = 1e-8;
int main() {
  istream &fin = cin;
  ostream &fout = cout;
  FILE *fpin = stdin;
  FILE *fpout = stdout;
  string s;
  fin >> s;
  bool p = true;
  string res;
  for (int i = s.size() - 1; i >= 0; i--) {
    if (p && s[i] == '/') continue;
    if (s[i] != '/')
      p = false;
    else
      p = true;
    res.push_back(s[i]);
  }
  reverse(res.begin(), res.end());
  if (res.empty()) res = "/";
  fout << res << endl;
}

```