#include <bits/stdc++.h>
using namespace std;
int main() {
  string s, r;
  cin >> s;
  int f = 0, i;
  for (char a : s) {
    if (a != '/' || f == 0) {
      r += a;
      (a == '/') ? f = 1 : f = 0;
    }
  }
  for (i = 0; i < r.size() - 1; i++) {
    cout << r[i];
  }
  if (r[r.size() - 1] != '/' || r.size() == 1) {
    cout << r[r.size() - 1];
  }
  cout << endl;
  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 s, r;
  cin >> s;
  int f = 0, i;
  for (char a : s) {
    if (a != '/' || f == 0) {
      r += a;
      (a == '/') ? f = 1 : f = 0;
    }
  }
  for (i = 0; i < r.size() - 1; i++) {
    cout << r[i];
  }
  if (r[r.size() - 1] != '/' || r.size() == 1) {
    cout << r[r.size() - 1];
  }
  cout << endl;
  return 0;
}

```