#include <bits/stdc++.h>
using namespace std;
string s;
int l;
bool flag = 0;
int main() {
  cin >> s;
  l = s.size();
  while (s[l - 1] == '/' && l >= 1) l--;
  if (!l) cout << "/" << endl;
  if (l == 1) cout << s[0] << endl;
  for (register int i = 0; i < l; i++) {
    if (flag) {
      if (s[i] != '/') {
        cout << s[i];
        flag = 0;
      }
    } else {
      cout << s[i];
      if (s[i] == '/') flag = 1;
    }
  }
  return 0;
}

### 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 s;
int l;
bool flag = 0;
int main() {
  cin >> s;
  l = s.size();
  while (s[l - 1] == '/' && l >= 1) l--;
  if (!l) cout << "/" << endl;
  if (l == 1) cout << s[0] << endl;
  for (register int i = 0; i < l; i++) {
    if (flag) {
      if (s[i] != '/') {
        cout << s[i];
        flag = 0;
      }
    } else {
      cout << s[i];
      if (s[i] == '/') flag = 1;
    }
  }
  return 0;
}

```