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

### 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;
int len, n;
string s;
int main() {
  cin >> s;
  len = s.length();
  n = 0;
  for (int i = 1; i < len; i++)
    if (s[i] != s[i - 1] || s[i] != '/') {
      ++n;
      cout << s[i - 1];
    }
  if (s[len - 1] != '/' || n == 0) cout << s[len - 1];
}

```