#include <bits/stdc++.h>
using namespace std;
string x;
int main() {
  cin >> x;
  cout << x[0];
  int st = 0;
  while (x[st] == x[st + 1]) {
    st++;
    if (st == x.size() - 1) break;
  }
  for (int i = st + 1; i < x.size(); i++) {
    if (x[i] != '/') {
      cout << x[i];
      continue;
    }
    while (x[i] == x[i + 1]) {
      i++;
      if (i == x.size() - 1) break;
    }
    if (i != x.size() - 1) cout << x[i];
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 x;
int main() {
  cin >> x;
  cout << x[0];
  int st = 0;
  while (x[st] == x[st + 1]) {
    st++;
    if (st == x.size() - 1) break;
  }
  for (int i = st + 1; i < x.size(); i++) {
    if (x[i] != '/') {
      cout << x[i];
      continue;
    }
    while (x[i] == x[i + 1]) {
      i++;
      if (i == x.size() - 1) break;
    }
    if (i != x.size() - 1) cout << x[i];
  }
}

```