#include <bits/stdc++.h>
using namespace std;
int big, small;
int sum;
int main() {
  string s;
  cin >> s;
  int len = s.size();
  while (s[len - 1] == '/' && len >= 1) len--;
  if (!len) cout << "/" << endl;
  if (len == 1) cout << s[0] << endl;
  for (int i = 0; i <= len - 1; i++) {
    if (s[i] == '/' && s[i + 1] != '/') cout << s[i];
    if (s[i] != '/') cout << s[i];
  }
}

### 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 big, small;
int sum;
int main() {
  string s;
  cin >> s;
  int len = s.size();
  while (s[len - 1] == '/' && len >= 1) len--;
  if (!len) cout << "/" << endl;
  if (len == 1) cout << s[0] << endl;
  for (int i = 0; i <= len - 1; i++) {
    if (s[i] == '/' && s[i + 1] != '/') cout << s[i];
    if (s[i] != '/') cout << s[i];
  }
}

```