#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  string str;
  cin >> str;
  str += '/';
  for (int i = 0; i < str.size(); i++) {
    if (str[i] == '/' and str[i + 1] == '/') {
      str.erase(str.begin() + i + 1);
      i--;
    }
  }
  int cnt = 0;
  for (int i = 0; i < str.size(); i++) {
    if (str[i] == '/') cnt++;
  }
  if (cnt != 1) str.erase(str.end() - 1);
  cout << str << endl;
}

### Prompt

Generate 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  string str;
  cin >> str;
  str += '/';
  for (int i = 0; i < str.size(); i++) {
    if (str[i] == '/' and str[i + 1] == '/') {
      str.erase(str.begin() + i + 1);
      i--;
    }
  }
  int cnt = 0;
  for (int i = 0; i < str.size(); i++) {
    if (str[i] == '/') cnt++;
  }
  if (cnt != 1) str.erase(str.end() - 1);
  cout << str << endl;
}

```