#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  cin >> s;
  string o = "";
  int n = s.length();
  if (n == 1 && s[0] == '/')
    cout << "/" << endl;
  else {
    for (int i = 0; i < s.length(); i++) {
      if (i != n - 1 && s[i] == '/' && s[i + 1] == '/')
        continue;
      else
        o += s[i];
    }
    if (o.length() > 1 && o[o.length() - 1] == '/') {
      for (int i = 0; i < o.length() - 1; i++) cout << o[i];
    } else
      cout << o << endl;
  }
}

### 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() {
  string s;
  cin >> s;
  string o = "";
  int n = s.length();
  if (n == 1 && s[0] == '/')
    cout << "/" << endl;
  else {
    for (int i = 0; i < s.length(); i++) {
      if (i != n - 1 && s[i] == '/' && s[i + 1] == '/')
        continue;
      else
        o += s[i];
    }
    if (o.length() > 1 && o[o.length() - 1] == '/') {
      for (int i = 0; i < o.length() - 1; i++) cout << o[i];
    } else
      cout << o << endl;
  }
}

```