#include <bits/stdc++.h>
using namespace std;
void solve() {
  string str;
  cin >> str;
  string temp;
  for (long long i = 0; i < str.size(); i++) {
    if (i == 0) temp += str[i];
    if (str[i] == '/' and temp[temp.size() - 1] == '/') continue;
    temp += str[i];
  }
  if (temp[temp.size() - 1] == '/' and temp.size() > 1)
    temp = temp.substr(0, temp.size() - 1);
  cout << temp;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}

### 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;
void solve() {
  string str;
  cin >> str;
  string temp;
  for (long long i = 0; i < str.size(); i++) {
    if (i == 0) temp += str[i];
    if (str[i] == '/' and temp[temp.size() - 1] == '/') continue;
    temp += str[i];
  }
  if (temp[temp.size() - 1] == '/' and temp.size() > 1)
    temp = temp.substr(0, temp.size() - 1);
  cout << temp;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}

```