#include <bits/stdc++.h>
using namespace std;
int main() {
  string s, s1;
  int ans = 0, i;
  vector<char> v;
  cin >> s;
  for (i = 0; i < s.size(); i++) {
    if (i == 0)
      v.push_back(s[i]);
    else {
      if (!(s[i] == '/' && v[v.size() - 1] == '/')) v.push_back(s[i]);
    }
  }
  if (v.size() > 1 && v[v.size() - 1] == '/') v.pop_back();
  for (i = 0; i < v.size(); i++) cout << v[i];
  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;
int main() {
  string s, s1;
  int ans = 0, i;
  vector<char> v;
  cin >> s;
  for (i = 0; i < s.size(); i++) {
    if (i == 0)
      v.push_back(s[i]);
    else {
      if (!(s[i] == '/' && v[v.size() - 1] == '/')) v.push_back(s[i]);
    }
  }
  if (v.size() > 1 && v[v.size() - 1] == '/') v.pop_back();
  for (i = 0; i < v.size(); i++) cout << v[i];
  return 0;
}

```