#include <bits/stdc++.h>
using namespace std;
int main() {
  string s1;
  cin >> s1;
  int i = 0;
  int g = s1.length();
  int k = 0;
  cout << s1[0];
  while (i < g) {
    if ((s1[i] == '/'))
      if (s1[i + 1] == '/') {
      } else {
        if (s1.length() - 1 == i) {
          if (s1[i] != '/') cout << s1[i];
        } else if (k != 0)
          cout << s1[i];
      }
    else {
      k++;
      cout << s1[i];
    }
    i++;
  }
  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;
int main() {
  string s1;
  cin >> s1;
  int i = 0;
  int g = s1.length();
  int k = 0;
  cout << s1[0];
  while (i < g) {
    if ((s1[i] == '/'))
      if (s1[i + 1] == '/') {
      } else {
        if (s1.length() - 1 == i) {
          if (s1[i] != '/') cout << s1[i];
        } else if (k != 0)
          cout << s1[i];
      }
    else {
      k++;
      cout << s1[i];
    }
    i++;
  }
  return 0;
}

```