#include <bits/stdc++.h>
using namespace std;
string s;
int n;
bool t;
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; ++i) {
    if (s[i] != '/') {
      t = 1;
      cout << "/";
      while (s[i] != '/' && i < n) {
        cout << s[i];
        ++i;
      }
    }
  }
  if (!t) cout << "/";
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
string s;
int n;
bool t;
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; ++i) {
    if (s[i] != '/') {
      t = 1;
      cout << "/";
      while (s[i] != '/' && i < n) {
        cout << s[i];
        ++i;
      }
    }
  }
  if (!t) cout << "/";
  return 0;
}

```