#include <bits/stdc++.h>
using namespace std;
bool ot[200];
int main() {
  string s;
  cin >> s;
  int n = s.length();
  int i = n - 1;
  while (s[n - 1] == '/' && n > 1) n--;
  for (int i = 0; i < n; i++)
    if (s[i] == '/') {
      i++;
      while (s[i] == '/') {
        ot[i] = 1;
        i++;
      }
    }
  for (int i = 0; i < n; i++)
    if (ot[i] == 0) cout << s[i];
}

### Prompt

Please create a solution in CPP 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;
bool ot[200];
int main() {
  string s;
  cin >> s;
  int n = s.length();
  int i = n - 1;
  while (s[n - 1] == '/' && n > 1) n--;
  for (int i = 0; i < n; i++)
    if (s[i] == '/') {
      i++;
      while (s[i] == '/') {
        ot[i] = 1;
        i++;
      }
    }
  for (int i = 0; i < n; i++)
    if (ot[i] == 0) cout << s[i];
}

```