#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
string s, ns;
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> s;
  ns += s[0];
  for (int i = 1; i < s.size(); i++) {
    if (s[i - 1] == '/' && s[i] == '/') continue;
    ns += s[i];
  }
  int j;
  for (j = ns.size() - 1; j >= 1; j--)
    if (ns[j] != '/') break;
  for (int i = 0; i <= j; i++) cout << ns[i];
  return EXIT_SUCCESS;
}

### 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;
const int N = 1e6 + 10;
string s, ns;
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> s;
  ns += s[0];
  for (int i = 1; i < s.size(); i++) {
    if (s[i - 1] == '/' && s[i] == '/') continue;
    ns += s[i];
  }
  int j;
  for (j = ns.size() - 1; j >= 1; j--)
    if (ns[j] != '/') break;
  for (int i = 0; i <= j; i++) cout << ns[i];
  return EXIT_SUCCESS;
}

```