#include <bits/stdc++.h>
using namespace std;
int main() {
  string a, b = "", c;
  cin >> a;
  int i = 0;
  for (i = 0; i < (a.size() - 1); i++) {
    if (a[i] == '/' && a[i + 1] == '/') continue;
    b += a[i];
  }
  if (a[i] != '/') b += a[i];
  if (b.size() == 0)
    cout << '/';
  else
    cout << b;
}

### Prompt

Your task is to create a Cpp solution 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;
int main() {
  string a, b = "", c;
  cin >> a;
  int i = 0;
  for (i = 0; i < (a.size() - 1); i++) {
    if (a[i] == '/' && a[i + 1] == '/') continue;
    b += a[i];
  }
  if (a[i] != '/') b += a[i];
  if (b.size() == 0)
    cout << '/';
  else
    cout << b;
}

```