#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  string a, b = "";
  cin >> a;
  b += a[0];
  for (int i = 1; i < a.size(); i++) {
    if (b[b.size() - 1] == a[i] && a[i] == '/') continue;
    b += a[i];
  }
  if (b.size() > 1 && b[b.size() - 1] == '/')
    cout << b.substr(0, b.size() - 1);
  else
    cout << b;
  return 0;
}

### Prompt

Your challenge is to write 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_base::sync_with_stdio(false);
  string a, b = "";
  cin >> a;
  b += a[0];
  for (int i = 1; i < a.size(); i++) {
    if (b[b.size() - 1] == a[i] && a[i] == '/') continue;
    b += a[i];
  }
  if (b.size() > 1 && b[b.size() - 1] == '/')
    cout << b.substr(0, b.size() - 1);
  else
    cout << b;
  return 0;
}

```