#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  cin >> s;
  int 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

Create a solution in CPP for 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() {
  string s;
  cin >> s;
  int 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];
}

```