#include <bits/stdc++.h>
int main() {
  std::string path;
  getline(std::cin, path);
  std::string output;
  bool flag(0);
  for (int k = 0; k < path.size(); k++) {
    if (path[k] != '/' || flag == 0) {
      output += path[k];
    }
    if (path[k] == '/') {
      flag = 1;
    } else {
      flag = 0;
    }
  }
  if (flag && output.size() > 1) {
    output = output.substr(0, output.size() - 1);
  }
  std::cout << output << std::endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
int main() {
  std::string path;
  getline(std::cin, path);
  std::string output;
  bool flag(0);
  for (int k = 0; k < path.size(); k++) {
    if (path[k] != '/' || flag == 0) {
      output += path[k];
    }
    if (path[k] == '/') {
      flag = 1;
    } else {
      flag = 0;
    }
  }
  if (flag && output.size() > 1) {
    output = output.substr(0, output.size() - 1);
  }
  std::cout << output << std::endl;
  return 0;
}

```